MTH120 College Algebra Chapter 2.5

 2.5 Quadratic Equations:

Quadratic equations are polynomial equations of the second degree, meaning the highest power of the variable (usually represented as $�$) is squared (${�}^2$). They have a general form:

$�{�}^2+��+�=0$

Where:

• $�$, $�$, and $�$ are constants, with $�$ not equal to 0.
• $�$ is the variable you're solving for.

Quadratic equations can have zero, one, or two real solutions (roots). The solutions can be found using the quadratic formula:

$�=\frac{-�\pm \sqrt{{�}^2-4��}}{2�}$

• The term inside the square root (${�}^2-4��$) is called the discriminant.
• If the discriminant is positive, there are two real solutions.
• If it's zero, there's one real solution (a repeated root).
• If it's negative, there are no real solutions (two complex roots).

To solve a quadratic equation, follow these steps:

1. Identify the coefficients $�$, $�$, and $�$.
2. Calculate the discriminant (${�}^2-4��$).
3. Use the quadratic formula to find the solutions for $�$.

For example, consider the quadratic equation $2{�}^2-5�+3=0$.

1. Here, $�=2$, $�=-5$, and $�=3$.
2. Calculate the discriminant: ${�}^2-4��=(-5{)}^2-4(2)(3)=25-24=1$.
3. Use the quadratic formula: $�=\frac{-(-5)\pm \sqrt{1}}{2(2)}$ Simplify: $�=\frac{5\pm 1}{4}$ This results in two solutions: ${�}_1=3\mathrm{/}2$ and ${�}_2=1\mathrm{/}2$.

So, the solutions to the equation $2{�}^2-5�+3=0$ are $�=\frac{3}{2}$ and $�=\frac{1}{2}$.

Solving quadratic equations by factoring involves rewriting the equation in factored form and then setting each factor equal to zero. Here's a step-by-step guide on how to do it:

1. Write the Equation in Standard Form: Make sure the quadratic equation is in the standard form: $�{�}^2+��+�=0$, where $�$, $�$, and $�$ are constants, and $�$ is not equal to zero.

2. Factor the Quadratic Expression: Factor the quadratic expression on the left side of the equation. You're looking for two expressions that multiply to the quadratic expression and add up to $�$.

3. Set Each Factor Equal to Zero: After factoring, you should have expressions on the left side, each equal to zero. Set each factor equal to zero and solve for $�$. These will be your potential solutions (roots).

4. Solve for $�$: Solve each of the equations obtained from step 3 for $�$. Depending on the factors, you may need to use different factoring techniques or the quadratic formula if factoring is not straightforward.

5. Check Your Solutions: Always check the solutions by plugging them back into the original equation to ensure they satisfy it.

Let's illustrate this process with an example:

Example: Solve the quadratic equation ${�}^2-5�+6=0$ by factoring.

1. The equation is already in standard form: ${�}^2-5�+6=0$.

2. Factor the quadratic expression on the left side: $(�-2)(�-3)=0$

3. Set each factor equal to zero: $�-2=0$ and $�-3=0$

4. Solve for $�$ in each equation:

   • $�-2=0$ gives $�=2$
   • $�-3=0$ gives $�=3$

5. Check your solutions: Plug $�=2$ and $�=3$ back into the original equation:

   • For $�=2$, you get $2^2-5(2)+6=0$, which is true.
   • For $�=3$, you get $3^2-5(3)+6=0$, which is also true.

So, the solutions to the equation ${�}^2-5�+6=0$ are $�=2$ and $�=3$.

The Zero-Product Property is a fundamental concept when dealing with quadratic equations. It states that if the product of two factors is equal to zero, then at least one of the factors must also be equal to zero.

Mathematically, if you have two expressions $�$ and $�$, and their product is zero ($��=0$), then either $�$ is equal to zero ($�=0$), or $�$ is equal to zero ($�=0$), or both.

The Zero-Product Property is extremely useful when solving quadratic equations because it allows us to break down a quadratic equation into simpler equations that can be easily solved.

Here's how you can use the Zero-Product Property to solve quadratic equations:

1. First, set the given quadratic equation equal to zero in standard form: $�{�}^2+��+�=0$.

2. Factor the quadratic expression on the left side if possible.

3. Once you have factored it, set each factor equal to zero using the Zero-Product Property.

4. Solve each equation obtained from step 3 for $�$. These are potential solutions (roots) of the quadratic equation.

5. Check your solutions by substituting them back into the original equation to ensure they satisfy it.

Here's an example to illustrate how the Zero-Product Property is used to solve a quadratic equation:

Example: Solve the quadratic equation ${�}^2-4�-5=0$ using the Zero-Product Property.

1. The equation is already in standard form: ${�}^2-4�-5=0$.

2. Factor the quadratic expression on the left side: $(�-5)(�+1)=0$.

3. Set each factor equal to zero using the Zero-Product Property:

   • $�-5=0$ and $�+1=0$.

4. Solve for $�$ in each equation:

   • $�-5=0$ gives $�=5$.
   • $�+1=0$ gives $�=-1$.

5. Check your solutions by substituting them back into the original equation:

   • For $�=5$, you get $5^2-4(5)-5=25-20-5=0$, which is true.
   • For $�=-1$, you get $(-1{)}^2-4(-1)-5=1+4-5=0$, which is also true.

So, the solutions to the equation ${�}^2-4�-5=0$ are $�=5$ and $�=-1$, and they are found using the Zero-Product Property.

Solving quadratic equations with a leading coefficient of 1 (i.e., quadratic equations in the form ${�}^2+��+�=0$) can be done using several methods, including factoring, completing the square, or using the quadratic formula. Here, I'll walk you through solving such equations using these methods.

Method 1: Factoring

1. Start with a quadratic equation in the form ${�}^2+��+�=0$.

2. Try to factor the left side of the equation into two binomials that multiply to zero. In other words, find two numbers that add up to $�$ and multiply to $�$.

3. Once you've factored it, set each factor equal to zero and solve for $�$.

4. Check your solutions by substituting them back into the original equation to ensure they satisfy it.

Method 2: Completing the Square

1. Start with a quadratic equation in the form ${�}^2+��+�=0$.

2. If necessary, move the constant term ($�$) to the other side of the equation so that the right side becomes $�$ (i.e., ${�}^2+��=-�$).

3. Add and subtract $(�\mathrm{/}2{)}^2$ inside the left side of the equation to create a perfect square trinomial. This involves adding $(�\mathrm{/}2{)}^2$ to both sides of the equation.

4. Write the left side of the equation as a square of a binomial. This should be $(�+(�\mathrm{/}2){)}^2$.

5. Take the square root of both sides.

6. Solve for $�$.

Method 3: Quadratic Formula

1. Start with a quadratic equation in the form ${�}^2+��+�=0$.

2. Identify the coefficients $�$, $�$, and $�$.

3. Use the quadratic formula to find the solutions for $�$:

   $�=\frac{-�\pm \sqrt{{�}^2-4��}}{2�}$

   • Calculate the discriminant (${�}^2-4��$) and determine its value.
   • If the discriminant is positive, there are two real solutions.
   • If it's zero, there's one real solution (a repeated root).
   • If it's negative, there are no real solutions (two complex roots).

4. Simplify and calculate the values of $�$.

5. Check your solutions by substituting them back into the original equation to ensure they satisfy it.

Here's an example using the quadratic formula:

Example: Solve the quadratic equation ${�}^2-4�+4=0$.

1. Identify the coefficients: $�=1$, $�=-4$, and $�=4$.

2. Use the quadratic formula:

   $�=\frac{-(-4)\pm \sqrt{(-4{)}^2-4(1)(4)}}{2(1)}$

3. Calculate the discriminant: ${�}^2-4��=(-4{)}^2-4(1)(4)=16-16=0$.

4. Since the discriminant is zero, there's one real solution.

5. Apply the quadratic formula:

   $�=\frac{4\pm 0}{2}$

6. Simplify:

   • ${�}_1=\frac{4}{2}=2$

So, the solution to the equation ${�}^2-4�+4=0$ is $�=2$, and it's a repeated root.

Sure, factoring a quadratic equation with a leading coefficient of 1 (i.e., in the form ${�}^2+��+�=0$) is a common method for solving such equations. Here are the steps to factor a quadratic equation:

Step 1: Start with the quadratic equation in the form ${�}^2+��+�=0$.

Step 2: Try to factor the left side of the equation into two binomials that multiply to zero. You're looking for two numbers that add up to $�$ (the coefficient of $�$) and multiply to $�$ (the constant term).

Step 3: Once you've factored it, set each factor equal to zero and solve for $�$.

Step 4: Check your solutions by substituting them back into the original equation to ensure they satisfy it.

Let's illustrate this process with an example:

Example: Factor the quadratic equation ${�}^2+5�+6=0$.

Step 1: The equation is already in standard form: ${�}^2+5�+6=0$.

Step 2: Factor the left side of the equation:

• You're looking for two numbers that add up to 5 (the coefficient of $�$) and multiply to 6 (the constant term).
• These two numbers are 2 and 3 because $2+3=5$ and $2\cdot 3=6$.

So, you can factor the equation as $(�+2)(�+3)=0$.

Step 3: Set each factor equal to zero:

• $�+2=0$
• $�+3=0$

Step 4: Solve for $�$ in each equation:

• $�+2=0$ gives $�=-2$
• $�+3=0$ gives $�=-3$

Step 5: Check your solutions:

• For $�=-2$, you get $(-2{)}^2+5(-2)+6=4-10+6=0$, which is true.
• For $�=-3$, you get $(-3{)}^2+5(-3)+6=9-15+6=0$, which is also true.

So, the solutions to the equation ${�}^2+5�+6=0$ are $�=-2$ and $�=-3$, and they were found by factoring the quadratic equation.

Factoring and solving a quadratic equation with a leading coefficient of 1 (i.e., in the form ${�}^2+��+�=0$) can be done using the factoring method. Here's how to do it step by step:

Step 1: Start with the quadratic equation in the form ${�}^2+��+�=0$.

Step 2: Try to factor the left side of the equation into two binomials that multiply to zero. Look for two numbers that add up to $�$ (the coefficient of $�$) and multiply to $�$ (the constant term).

Step 3: Once you've factored it, set each factor equal to zero and solve for $�$.

Step 4: Check your solutions by substituting them back into the original equation to ensure they satisfy it.

Let's use an example to illustrate the process:

Example: Factor and solve the quadratic equation ${�}^2+7�+10=0$.

Step 1: The equation is already in standard form: ${�}^2+7�+10=0$.

Step 2: Factor the left side of the equation:

• You're looking for two numbers that add up to 7 (the coefficient of $�$) and multiply to 10 (the constant term).
• These two numbers are 2 and 5 because $2+5=7$ and $2\cdot 5=10$.

So, you can factor the equation as $(�+2)(�+5)=0$.

Step 3: Set each factor equal to zero:

• $�+2=0$
• $�+5=0$

Step 4: Solve for $�$ in each equation:

• $�+2=0$ gives $�=-2$
• $�+5=0$ gives $�=-5$

Step 5: Check your solutions:

• For $�=-2$, you get $(-2{)}^2+7(-2)+10=4-14+10=0$, which is true.
• For $�=-5$, you get $(-5{)}^2+7(-5)+10=25-35+10=0$, which is also true.

So, the solutions to the equation ${�}^2+7�+10=0$ are $�=-2$ and $�=-5$, and they were found by factoring the quadratic equation.

The Zero-Product Property is a valuable tool for solving quadratic equations, even when they are written as the difference of squares. Here's how you can use the Zero-Product Property to solve a quadratic equation in the form of the difference of squares:

Step 1: Start with a quadratic equation in the difference of squares form: ${�}^2-{�}^2=0$.

Step 2: Recognize that this equation can be factored as $(�+�)(�-�)=0$. This is based on the formula for the difference of squares, which states that ${�}^2-{�}^2=(�+�)(�-�)$.

Step 3: Use the Zero-Product Property, which states that if the product of two factors is equal to zero, then at least one of the factors must also be equal to zero.

So, set each factor equal to zero and solve for $�$:

1. $�+�=0$
2. $�-�=0$

Step 4: Solve each equation separately for $�$:

1. $�+�=0$ gives $�=-�$.
2. $�-�=0$ gives $�=�$.

Step 5: Check your solutions by substituting them back into the original equation (${�}^2-{�}^2=0$) to ensure they satisfy it.

Example: Solve the equation ${�}^2-9=0$ using the Zero-Product Property.

Step 1: The equation is already in the difference of squares form: ${�}^2-9=0$.

Step 2: Recognize that this can be factored as $(�+3)(�-3)=0$ based on the difference of squares.

Step 3: Use the Zero-Product Property and set each factor equal to zero:

1. $�+3=0$
2. $�-3=0$

Step 4: Solve each equation separately for $�$:

1. $�+3=0$ gives $�=-3$.
2. $�-3=0$ gives $�=3$.

Step 5: Check your solutions by substituting them back into the original equation:

• For $�=-3$, you get $(-3{)}^2-9=9-9=0$, which is true.
• For $�=3$, you get $3^2-9=9-9=0$, which is also true.

So, the solutions to the equation ${�}^2-9=0$ are $�=-3$ and $�=3$, found using the Zero-Product Property and factoring the difference of squares.

To solve a quadratic equation by factoring when the leading coefficient is not 1 (i.e., the equation is in the form $�{�}^2+��+�=0$), you can use the following steps:

Step 1: Write the quadratic equation in standard form: $�{�}^2+��+�=0$, where $�$, $�$, and $�$ are constants and $�$ is not equal to 0.

Step 2: Try to factor the left side of the equation into two binomials. Look for two numbers that multiply to $�\cdot �$ (the product of the leading coefficient $�$ and the constant term $�$) and add up to $�$ (the coefficient of $�$).

Step 3: Once you've factored the left side, set each factor equal to zero and solve for $�$.

Step 4: Check your solutions by substituting them back into the original equation to ensure they satisfy it.

Here's an example to illustrate the process:

Example: Solve the quadratic equation $2{�}^2-7�-3=0$ by factoring.

Step 1: The equation is already in standard form: $2{�}^2-7�-3=0$.

Step 2: Factor the left side of the equation:

• You're looking for two numbers that multiply to $2\cdot (-3)=-6$ and add up to $-7$ (the coefficient of $�$).
• These two numbers are $-9$ and $2$ because $-9+2=-7$ and $-9\cdot 2=-18$.

So, you can factor the equation as $2{�}^2-9�+2�-3=0$, and then group the terms:

$2�(�-3)+1(�-3)=0$

Now, you have a common factor of $(�-3)$:

$(2�+1)(�-3)=0$

Step 3: Set each factor equal to zero:

• $2�+1=0$
• $�-3=0$

Step 4: Solve for $�$ in each equation:

1. $2�+1=0$ gives $2�=-1$ and $�=-1\mathrm{/}2$.
2. $�-3=0$ gives $�=3$.

Step 5: Check your solutions:

• For $�=-1\mathrm{/}2$, you get $2(-1\mathrm{/}2{)}^2-7(-1\mathrm{/}2)-3=1\mathrm{/}2+7\mathrm{/}2-3=0$, which is true.
• For $�=3$, you get $2(3{)}^2-7(3)-3=18-21-3=0$, which is also true.

So, the solutions to the equation $2{�}^2-7�-3=0$ are $�=-1\mathrm{/}2$ and $�=3$, and they were found by factoring the quadratic equation.

Solving a quadratic equation using the method of grouping involves splitting the middle term of the quadratic expression and then factoring by grouping. Here's how you can do it step by step:

Step 1: Start with a quadratic equation in standard form: $�{�}^2+��+�=0$, where $�$, $�$, and $�$ are constants, and $�$ is not equal to 0.

Step 2: Look for two numbers $�$ and $�$ such that $�\cdot �=�\cdot �$ (the product of the leading coefficient $�$ and the constant term $�$), and $�+�=�$ (the coefficient of $�$).

Step 3: Rewrite the middle term ($��$) using the values of $�$ and $�$. You can do this by splitting $��$ into $��$ and $��$. So, your quadratic equation becomes: $�{�}^2+��+��+�=0$.

Step 4: Group the terms in pairs and factor by grouping. You will have two groups: $(�{�}^2+��)$ and $(��+�)$.

Step 5: Factor out the greatest common factor (GCF) from each group.

Step 6: The factored expressions from each group should be the same. Set them equal to each other and solve for $�$.

Step 7: Check your solutions by substituting them back into the original equation to ensure they satisfy it.

Here's an example to illustrate the process:

Example: Solve the quadratic equation ${�}^2+3�-4=0$ using the method of grouping.

Step 1: The equation is already in standard form: ${�}^2+3�-4=0$.

Step 2: Find two numbers $�$ and $�$ such that $�\cdot �=1\cdot (-4)=-4$ and $�+�=3$.

The numbers that fit these criteria are 4 and -1, because $4\cdot (-1)=-4$ and $4+(-1)=3$.

Step 3: Rewrite the middle term using $�$ and $�$:

${�}^2+4�-�-4=0$

Step 4: Group the terms in pairs and factor by grouping:

$({�}^2+4�)-(�+4)=0$

Step 5: Factor out the GCF from each group:

$�(�+4)-1(�+4)=0$

Step 6: Set the factored expressions from each group equal to each other:

$�(�+4)-1(�+4)=0$

Step 7: Solve for $�$:

• $�(�+4)-1(�+4)=0$ simplifies to $(�-1)(�+4)=0$.
• Setting each factor equal to zero: $�-1=0$ gives $�=1$, and $�+4=0$ gives $�=-4$.

Step 8: Check your solutions:

• For $�=1$, you get $1^2+3(1)-4=1+3-4=0$, which is true.
• For $�=-4$, you get $(-4{)}^2+3(-4)-4=16-12-4=0$, which is also true.

So, the solutions to the equation ${�}^2+3�-4=0$ are $�=1$ and $�=-4$, found by factoring using the method of grouping.

Solving a polynomial equation of higher degree by factoring can be a bit more challenging than solving quadratic equations, but it's still possible using a similar approach. Here are the general steps to solve a polynomial equation of higher degree by factoring:

Step 1: Write the polynomial equation in standard form. Standard form means the polynomial is set equal to zero, and it should be ordered from highest degree term to lowest degree term.

Step 2: Factor out the greatest common factor (GCF) if there is one. This can often simplify the equation.

Step 3: Try to factor the remaining polynomial into a product of lower-degree polynomials. Look for patterns or grouping that may help you factor it.

Step 4: Set each factor equal to zero and solve for the variable(s). This will give you a set of potential solutions.

Step 5: Check your solutions by substituting them back into the original equation to ensure they satisfy it.

Here's an example to illustrate the process:

Example: Solve the polynomial equation ${�}^3-8{�}^2+15�-6=0$ by factoring.

Step 1: The equation is already in standard form: ${�}^3-8{�}^2+15�-6=0$.

Step 2: There's no GCF to factor out.

Step 3: Try to factor the polynomial:

• Notice that $�=1$ is a root (a solution) because $1^3-8(1{)}^2+15(1)-6=1-8+15-6=2-2=0$.
• You can use synthetic division or polynomial long division to divide ${�}^3-8{�}^2+15�-6$ by $(�-1)$. This will give you a quadratic quotient: ${�}^2-7�+6$.

Now, we need to factor the quadratic expression ${�}^2-7�+6$:

$(�-1)({�}^2-7�+6)=0$

Step 4: Set each factor equal to zero:

1. $�-1=0$ gives $�=1$.
2. ${�}^2-7�+6=0$

Now, focus on solving the quadratic equation ${�}^2-7�+6=0$. This quadratic factors as $(�-6)(�-1)=0$.

Step 5: Set each factor equal to zero:

1. $�-6=0$ gives $�=6$.
2. $�-1=0$ gives $�=1$.

Step 6: Check your solutions:

• For $�=1$, you've already verified that it satisfies the original equation.
• For $�=6$, substitute it into the original equation:

$$<span\; style="background-color:\; white;">(</span>\mathrm{<span\; style="background-color:\; white;">6</span>}{<span\; style="background-color:\; white;">)</span>}^{\mathrm{<span\; style="background-color:\; white;">3</span>}}<span\; style="background-color:\; white;">-</span>\mathrm{<span\; style="background-color:\; white;">8</span>}<span\; style="background-color:\; white;">(</span>\mathrm{<span\; style="background-color:\; white;">6</span>}{<span\; style="background-color:\; white;">)</span>}^{\mathrm{<span\; style="background-color:\; white;">2</span>}}<span\; style="background-color:\; white;">+</span>\mathrm{<span\; style="background-color:\; white;">15</span>}<span\; style="background-color:\; white;">(</span>\mathrm{<span\; style="background-color:\; white;">6</span>}<span\; style="background-color:\; white;">)</span><span\; style="background-color:\; white;">-</span>\mathrm{<span\; style="background-color:\; white;">6</span>}<span\; style="background-color:\; white;">=</span>\mathrm{<span\; style="background-color:\; white;">216</span>}<span\; style="background-color:\; white;">-</span>\mathrm{<span\; style="background-color:\; white;">288</span>}<span\; style="background-color:\; white;">+</span>\mathrm{<span\; style="background-color:\; white;">90</span>}<span\; style="background-color:\; white;">-</span>\mathrm{<span\; style="background-color:\; white;">6</span>}<span\; style="background-color:\; white;">=</span>\mathrm{<span\; style="background-color:\; white;">0</span>}$$

So, the solutions to the polynomial equation ${�}^3-8{�}^2+15�-6=0$ are $�=1$ and $�=6$, found by factoring and checking the roots.

The Square Root Property is a useful method for solving quadratic equations in the form ${�}^2=�$, where $�$ is a non-negative constant. The Square Root Property states that if ${�}^2=�$, then $�$ is equal to the positive and negative square roots of $�$.

Here are the steps to solve a quadratic equation using the Square Root Property:

Step 1: Start with a quadratic equation in the form ${�}^2=�$, where $�$ is a non-negative constant.

Step 2: Take the square root of both sides of the equation.

Step 3: Include both the positive and negative square roots on one side of the equation.

Step 4: Solve for $�$.

Step 5: Check your solutions by substituting them back into the original equation to ensure they satisfy it.

Here's an example to illustrate the process:

Example: Solve the quadratic equation ${�}^2=25$ using the Square Root Property.

Step 1: The equation is already in the form ${�}^2=�$, where $�=25$.

Step 2: Take the square root of both sides:

$\sqrt{{�}^2}=\sqrt{25}$

Step 3: Include both the positive and negative square roots:

$�=\pm \sqrt{25}$

Step 4: Solve for $�$:

1. $�=\sqrt{25}$ gives $�=5$.
2. $�=-\sqrt{25}$ gives $�=-5$.

Step 5: Check your solutions:

• For $�=5$, you get $5^2=25$, which is true.
• For $�=-5$, you get $(-5{)}^2=25$, which is also true.

So, the solutions to the equation ${�}^2=25$ are $�=5$ and $�=-5$, found using the Square Root Property.

If you have a quadratic equation with an ${�}^2$ term but no $�$ term, it will have the form ${�}^2=�$, where $�$ is a constant. You can use the Square Root Property to solve it as follows:

Step 1: Start with the quadratic equation in the form ${�}^2=�$.

Step 2: Take the square root of both sides of the equation.

Step 3: Include both the positive and negative square roots on one side of the equation.

Step 4: Solve for $�$.

Step 5: Check your solutions by substituting them back into the original equation to ensure they satisfy it.

Here's an example to illustrate the process:

Example: Solve the quadratic equation ${�}^2=49$ using the Square Root Property.

Step 1: The equation is already in the form ${�}^2=�$, where $�=49$.

Step 2: Take the square root of both sides:

$\sqrt{{�}^2}=\sqrt{49}$

Step 3: Include both the positive and negative square roots:

$�=\pm \sqrt{49}$

Step 4: Solve for $�$:

1. $�=\sqrt{49}$ gives $�=7$.
2. $�=-\sqrt{49}$ gives $�=-7$.

Step 5: Check your solutions:

• For $�=7$, you get $7^2=49$, which is true.
• For $�=-7$, you get $(-7{)}^2=49$, which is also true.

So, the solutions to the equation ${�}^2=49$ are $�=7$ and $�=-7$, found using the Square Root Property.

Let's solve a simple quadratic equation using the Square Root Property step by step.

Example: Solve the quadratic equation ${�}^2=16$ using the Square Root Property.

Step 1: Start with the quadratic equation in the form ${�}^2=�$, where $�$ is a constant. In this case, $�=16$.

Step 2: Take the square root of both sides of the equation:

$\sqrt{{�}^2}=\sqrt{16}$

Step 3: Include both the positive and negative square roots:

$�=\pm \sqrt{16}$

Step 4: Solve for $�$:

1. $�=\sqrt{16}$ gives $�=4$.
2. $�=-\sqrt{16}$ gives $�=-4$.

Step 5: Check your solutions:

• For $�=4$, you get $4^2=16$, which is true.
• For $�=-4$, you get $(-4{)}^2=16$, which is also true.

So, the solutions to the equation ${�}^2=16$ are $�=4$ and $�=-4$, found using the Square Root Property.

Completing the square is a method used to solve quadratic equations in the form $�{�}^2+��+�=0$. Here are the general steps to complete the square and solve a quadratic equation:

Step 1: Start with a quadratic equation in the form $�{�}^2+��+�=0$, where $�$, $�$, and $�$ are constants and $�$ is not equal to 0.

Step 2: If $�$ is not equal to 1, divide the entire equation by $�$ to make the coefficient of ${�}^2$ equal to 1. This step simplifies the equation.

Step 3: Move the constant term ($�$) to the other side of the equation by subtracting it from both sides, so that you have $�{�}^2+��=-�$.

Step 4: To complete the square, add and subtract $(�\mathrm{/}2�{)}^2$ on the left side of the equation. This term, $(�\mathrm{/}2�{)}^2$, is known as the "completing the square" term and helps in factoring a perfect square trinomial.

Step 5: Rewrite the left side of the equation as a perfect square trinomial. The goal is to have it in the form $(�+�{)}^2$, where $�$ is a constant.

Step 6: Take the square root of both sides to solve for $�$.

Step 7: Simplify and solve for $�$.

Step 8: Check your solution by substituting it back into the original equation to ensure it satisfies it.

Here's an example to illustrate the process:

Example: Solve the quadratic equation ${�}^2-4�-5=0$ by completing the square.

Step 1: The equation is already in the form $�{�}^2+��+�=0$ with $�=1$, so we can proceed.

Step 2: No need to divide by $�$ since $�=1$.

Step 3: Move the constant term ($-5$) to the other side:

${�}^2-4�=5$

Step 4: Add and subtract $(4\mathrm{/}2\cdot 1{)}^2=4$ on the left side:

${�}^2-4�+4-4=5$

Step 5: Rewrite the left side as a perfect square trinomial:

$(�-2{)}^2-4=5$

Step 6: Take the square root of both sides:

$\sqrt{(�-2{)}^2-4}=\pm \sqrt{5}$

Step 7: Simplify and solve for $�$:

$�-2=\pm \sqrt{5}$

For $�-2=\sqrt{5}$, add 2 to both sides: $�=2+\sqrt{5}$.

For $�-2=-\sqrt{5}$, add 2 to both sides: $�=2-\sqrt{5}$.

Step 8: Check your solutions:

• For $�=2+\sqrt{5}$, substitute it into the original equation: $(2+\sqrt{5}{)}^2-4(2+\sqrt{5})-5=0$, which is true.
• For $�=2-\sqrt{5}$, substitute it into the original equation: $(2-\sqrt{5}{)}^2-4(2-\sqrt{5})-5=0$, which is also true.

So, the solutions to the equation ${�}^2-4�-5=0$ are $�=2+\sqrt{5}$ and $�=2-\sqrt{5}$, found by completing the square.

The quadratic formula is a powerful method for solving quadratic equations of the form $�{�}^2+��+�=0$, where $�$, $�$, and $�$ are constants and $�$ is not equal to 0. The quadratic formula is:

$�=\frac{-�\pm \sqrt{{�}^2-4��}}{2�}$

Here are the steps to use the quadratic formula to solve a quadratic equation:

Step 1: Start with a quadratic equation in the form $�{�}^2+��+�=0$.

Step 2: Identify the coefficients $�$, $�$, and $�$.

Step 3: Use the quadratic formula to find the values of $�$. Be sure to calculate both the plus and minus versions of the formula to find both solutions, if they exist.

Step 4: Simplify and calculate the values of $�$:

• Calculate the discriminant ($�$) under the square root, which is ${�}^2-4��$.
• If $�>0$, there are two distinct real solutions for $�$.
• If $�=0$, there is one real solution for $�$.
• If $�<0$, there are no real solutions (the solutions are complex).

Step 5: Check your solutions by substituting them back into the original equation to ensure they satisfy it.

Here's an example to illustrate the process:

Example: Solve the quadratic equation $2{�}^2-5�+2=0$ using the quadratic formula.

Step 1: The equation is already in the form $�{�}^2+��+�=0$, where $�=2$, $�=-5$, and $�=2$.

Step 2: Identify the coefficients: $�=2$, $�=-5$, $�=2$.

Step 3: Use the quadratic formula:

$�=\frac{-(-5)\pm \sqrt{(-5{)}^2-4(2)(2)}}{2(2)}$

Step 4: Simplify and calculate:

• Calculate the discriminant: $�=(-5{)}^2-4(2)(2)=25-16=9$.

Since $�>0$, there are two distinct real solutions.

• Calculate both solutions:
  • ${�}_1=\frac{5+\sqrt{9}}{4}=\frac{5+3}{4}=\frac{8}{4}=2$
  • ${�}_2=\frac{5-\sqrt{9}}{4}=\frac{5-3}{4}=\frac{2}{4}=\frac{1}{2}$

Step 5: Check your solutions:

• For $�=2$, you get $2(2{)}^2-5(2)+2=8-10+2=0$, which is true.
• For $�=1\mathrm{/}2$, you get $2(1\mathrm{/}2{)}^2-5(1\mathrm{/}2)+2=1\mathrm{/}2-5\mathrm{/}2+2=0$, which is also true.

So, the solutions to the equation $2{�}^2-5�+2=0$ are $�=2$ and $�=1\mathrm{/}2$, found using the quadratic formula.

Certainly! Let's solve a quadratic equation using the quadratic formula. Here's a step-by-step example:

Example: Solve the quadratic equation $3{�}^2-6�+2=0$ using the quadratic formula.

Step 1: Start with the quadratic equation in the form $�{�}^2+��+�=0$.

Step 2: Identify the coefficients $�$, $�$, and $�$. In this case, $�=3$, $�=-6$, and $�=2$.

Step 3: Use the quadratic formula:

$�=\frac{-�\pm \sqrt{{�}^2-4��}}{2�}$

Plugging in the values:

$�=\frac{-(-6)\pm \sqrt{(-6{)}^2-4(3)(2)}}{2(3)}$

Step 4: Simplify and calculate:

• Calculate the discriminant ($�$) under the square root:

$�=(-6{)}^2-4(3)(2)=36-24=12$

Since $�>0$, there are two distinct real solutions.

• Calculate both solutions:

${�}_1=\frac{6+\sqrt{12}}{6}=\frac{6+2\sqrt{3}}{6}=\frac{3+\sqrt{3}}{3}$

${�}_2=\frac{6-\sqrt{12}}{6}=\frac{6-2\sqrt{3}}{6}=\frac{3-\sqrt{3}}{3}$

Step 5: Check your solutions:

• For $�=\frac{3+\sqrt{3}}{3}$, you get:

$3{\left(\frac{3+\sqrt{3}}{3}\right)}^2-6\left(\frac{3+\sqrt{3}}{3}\right)+2=3\left(\frac{12+6\sqrt{3}+3}{9}\right)-6\left(\frac{3+\sqrt{3}}{3}\right)+2=12+6\sqrt{3}+3-6(3+\sqrt{3})+2=0$

• For $�=\frac{3-\sqrt{3}}{3}$, you get:

$3{\left(\frac{3-\sqrt{3}}{3}\right)}^2-6\left(\frac{3-\sqrt{3}}{3}\right)+2=3\left(\frac{12-6\sqrt{3}+3}{9}\right)-6\left(\frac{3-\sqrt{3}}{3}\right)+2=12-6\sqrt{3}+3-6(3-\sqrt{3})+2=0$

So, the solutions to the equation $3{�}^2-6�+2=0$ are $�=\frac{3+\sqrt{3}}{3}$ and $�=\frac{3-\sqrt{3}}{3}$, found using the quadratic formula.

The discriminant ($�$) is a value that plays a crucial role in determining the nature of the solutions (roots) of a quadratic equation. It's used to assess whether the equation has real solutions, complex solutions, or repeated solutions. The discriminant is calculated using the coefficients of the quadratic equation $�{�}^2+��+�=0$, where $�$, $�$, and $�$ are constants.

The formula to calculate the discriminant ($�$) is as follows:

$�={�}^2-4��$

Here's what the discriminant tells us about the solutions of the quadratic equation:

1. If $�>0$: The equation has two distinct real solutions. This means that the graph of the quadratic equation intersects the x-axis at two different points.

2. If $�=0$: The equation has one real solution with multiplicity 2 (a repeated root). This means that the graph of the quadratic equation touches the x-axis at a single point (tangent).

3. If $�<0$: The equation has no real solutions. In this case, the solutions are complex conjugate pairs, and the graph of the quadratic equation does not intersect the x-axis.

The discriminant helps you determine the nature of the solutions before actually calculating them. It's a valuable tool when dealing with quadratic equations, as it provides insight into the behavior of the equation's roots.

The discriminant ($�$) in the quadratic formula $�=\frac{-�\pm \sqrt{�}}{2�}$ can be used to find the nature of the solutions (roots) to a quadratic equation. Here's how to determine the nature of the solutions using the discriminant:

1. If $�>0$: In this case, the discriminant is positive. The equation has two distinct real solutions. These solutions are real numbers, and the graph of the quadratic equation intersects the x-axis at two different points.

2. If $�=0$: When the discriminant is exactly zero, the equation has one real solution with multiplicity 2. This means that there is a repeated root. The graph of the quadratic equation touches (is tangent to) the x-axis at a single point.

3. If $�<0$: If the discriminant is negative, there are no real solutions to the equation. Instead, the solutions are complex conjugate pairs. Complex numbers have the form $�+��$, where $�$ and $�$ are real numbers, and $�$ is the imaginary unit. The graph of the quadratic equation does not intersect the x-axis in the real number plane.

Here are some examples to illustrate these cases:

Example 1: Solve the equation ${�}^2-4�+4=0$ and find the discriminant.

Solution: The equation is already in standard form, and we can use the quadratic formula:

$�=1$, $�=-4$, and $�=4$.

The discriminant is $�=(-4{)}^2-4(1)(4)=16-16=0$.

Since $�=0$, there is one real solution with multiplicity 2, which means there is a repeated root. The solution is $�=2$.

Example 2: Solve the equation ${�}^2+2�+5=0$ and find the discriminant.

Solution: Again, we use the quadratic formula:

$�=1$, $�=2$, and $�=5$.

The discriminant is $�=(2{)}^2-4(1)(5)=4-20=-16$.

Since $�<0$, there are no real solutions to this equation. The solutions are complex conjugate pairs. The solutions are $�=-1+3�$ and $�=-1-3�$, where $�$ is the imaginary unit.

By examining the discriminant, you can quickly determine the nature of the solutions to a quadratic equation without solving for the specific values of $�$.

The Pythagorean Theorem is a fundamental principle in geometry that relates the sides of a right triangle. It states that in a right triangle:

${�}^2+{�}^2={�}^2$

Where:

• $�$ and $�$ are the lengths of the two shorter sides (legs) of the right triangle.
• $�$ is the length of the longest side, which is the hypotenuse.

The Pythagorean Theorem allows you to calculate the length of one side of a right triangle if you know the lengths of the other two sides. It is widely used in various mathematical and real-world applications.

Here are some common uses of the Pythagorean Theorem:

1. Finding the Length of a Missing Side: Given a right triangle with two known side lengths, you can use the theorem to calculate the length of the third side.

2. Checking for Right Triangles: You can use the theorem to determine if a given triangle is a right triangle. If the theorem holds true for the given side lengths, it's a right triangle.

3. Distance Calculation: In coordinate geometry, you can use the Pythagorean Theorem to calculate the distance between two points in a Cartesian plane.

4. Construction: It's used in construction and engineering to ensure that structures are built with right angles.

5. Navigation: In navigation, it's used to calculate distances and angles, such as when finding the distance between two points on a map.

Here's an example of how to use the Pythagorean Theorem:

Example: Suppose you have a right triangle with one leg measuring 3 units and the other leg measuring 4 units. Find the length of the hypotenuse.

Solution: Use the Pythagorean Theorem:

${�}^2+{�}^2={�}^2$

Substitute the known values:

$3^2+4^2={�}^2$

Simplify:

$9+16={�}^2$

$25={�}^2$

Take the square root of both sides to find $�$:

$�=\sqrt{25}=5$

So, the length of the hypotenuse is 5 units.

Use the Pythagorean Theorem to solve the right triangle problem: Leg a measures 4 units, leg b measures 3 units. Find the length of the hypotenuse.To solve this right triangle problem and find the length of the hypotenuse (

$�$), you can use the Pythagorean Theorem, which states that ${�}^2+{�}^2={�}^2$ in a right triangle, where $�$ and $�$ are the lengths of the legs, and $�$ is the length of the hypotenuse.

In this case, $�=4$ units and $�=3$ units. Now, let's plug these values into the Pythagorean Theorem and solve for $�$:

${�}^2+{�}^2={�}^2$ $(4{)}^2+(3{)}^2={�}^2$ $16+9={�}^2$ $25={�}^2$

To find $�$, take the square root of both sides:

$�=\sqrt{25}$ $�=5$

So, the length of the hypotenuse ($�$) in this right triangle is 5 units.

To solve the quadratic equation $6{�}^2+7�-20=0$, you can use the quadratic formula:

$�=\frac{-�\pm \sqrt{{�}^2-4��}}{2�}$

In this equation, $�=6$, $�=7$, and $�=-20$. Plug these values into the formula:

$�=\frac{-7\pm \sqrt{7^2-4(6)(-20)}}{2(6)}$

Now, calculate the discriminant ($�$) under the square root:

$�=7^2-4(6)(-20)$ $�=49+480$ $�=529$

Since $�>0$, there are two distinct real solutions for $�$. Now, plug the values into the formula:

${�}_1=\frac{-7+\sqrt{529}}{2(6)}$ ${�}_2=\frac{-7-\sqrt{529}}{2(6)}$

Simplify:

${�}_1=\frac{-7+23}{12}$ ${�}_2=\frac{-7-23}{12}$

Now, calculate ${�}_1$ and ${�}_2$:

${�}_1=\frac{16}{12}=\frac{4}{3}$ ${�}_2=\frac{-30}{12}=-\frac{5}{2}$

So, the solutions to the equation $6{�}^2+7�-20=0$ are $�=\frac{4}{3}$ and $�=-\frac{5}{2}$.