Solve the given quadratic equation by factoring. 3 x 2 + 12 x = 0 3x^2+12x=0 3 x 2 + 12 x = 0

x = 0 , x = − 4 x=0,x=-4 x = 0 , x = − 4

x = 3 , x = 4 x=3,x=4 x = 3 , x = 4

x = − 3 , x = − 4 x=-3,x=-4 x = − 3 , x = − 4

x = 1 , x = 4 x=1,x=4 x = 1 , x = 4

Solve the given equation by factoring. 2 x 2 + 7 x + 6 = 0 2x^2+7x+6=0 2 x 2 + 7 x + 6 = 0

x = − 2 , x = − 3 2 x=-2,x=-\frac32 x = − 2 , x = − 2 3

x = 3 , x = 6 7 x=3,x=\frac67 x = 3 , x = 7 6

x = − 2 , x = 0 x=-2,x=0 x = − 2 , x = 0

x = 2 , x = 3 2 x=2,x=\frac32 x = 2 , x = 2 3

1. Equations and Inequalities

Intro to Quadratic Equations

1. Equations and Inequalities

Intro to Quadratic Equations: Videos & Practice Problems

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Topic summary

Quadratic equations, or second-degree polynomials, take the form $a x^{2} + b x + c = 0$ . To solve them, factor the equation and set each factor to zero, yielding two possible solutions for x. Understanding coefficients a, b, and c is crucial, as they determine the equation's characteristics and solutions. Mastering this process is essential for solving quadratic equations effectively.

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Introduction to Quadratic Equations

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Introduction to Quadratic Equations Video Summary

Quadratic equations are a fundamental concept in algebra, representing polynomial equations of degree 2. They can be recognized by their standard form, which is expressed as:

$ ax^2 + bx + c = 0 $

In this equation, $ a $, $ b $, and $ c $ are coefficients, where $ a $ is the coefficient of the $ x^2 $ term, $ b $ is the coefficient of the $ x $ term, and $ c $ is the constant term. The highest power of the variable $ x $ in a quadratic equation is 2, which is why it is also referred to as a second-degree polynomial.

To write a quadratic equation in standard form, all terms must be on one side of the equation, arranged in descending order of their powers. For example, if we start with the equation $ 5x^2 = x - 3 $, we can rearrange it by moving all terms to one side:

$ 5x^2 - x + 3 = 0 $

In this case, we identify the coefficients as follows: $ a = 5 $, $ b = -1 $ (since $ -x $ can be viewed as $ -1x $), and $ c = 3 $.

Another example is the equation $ -2x^2 + \frac{5}{3} = 0 $. This equation is already in standard form, and we can identify the coefficients: $ a = -2 $, $ b = 0 $ (since there is no $ x $ term), and $ c = \frac{5}{3} $. It is important to note that $ b $ and $ c $ can be zero or even fractions, as long as the $ x^2 $ term is present, confirming it as a quadratic equation.

Understanding how to manipulate and identify the components of quadratic equations is crucial for solving them, which will be explored in further lessons.

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Problem

Write the given quadratic equation in standard form. Identify a, b, and c. $-4x^2+x=8$

a = - 4, b = 0, c = - 8

a = - 4, b = 1, c = 8

a = - 4, b = 1, c = - 8

a = 2, b = 1, c = 0

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Solving Quadratic Equations by Factoring

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Solving Quadratic Equations by Factoring Video Summary

When solving quadratic equations, the goal is to find values for $ x $ that satisfy the equation, similar to solving linear equations. However, quadratic equations, which typically take the form $ ax^2 + bx + c = 0 $, often yield two solutions for $ x $. To solve these equations, we can utilize the method of factoring.

To begin, we need to express the quadratic equation in standard form, ensuring all terms are on one side and arranged in descending order of power. For example, if we have the equation $ x^2 - 9x = -20 $, we can add 20 to both sides to rewrite it as $ x^2 - 9x + 20 = 0 $.

Next, we factor the quadratic expression. In this case, we identify the coefficients $ a $, $ b $, and $ c $ from the standard form. Here, $ a = 1 $, $ b = -9 $, and $ c = 20 $. We then apply the $ ac $ method, which involves finding two numbers that multiply to $ ac $ (which is $ 20 $) and add to $ b $ (which is $ -9 $). The factors that meet these criteria are $ -4 $ and $ -5 $, since $ -4 \times -5 = 20 $ and $ -4 + -5 = -9 $.

With the factors identified, we can express the quadratic as $ (x - 4)(x - 5) = 0 $. To find the solutions for $ x $, we set each factor equal to zero:

1. $ x - 4 = 0 $ leads to $ x = 4 $

2. $ x - 5 = 0 $ leads to $ x = 5 $

Thus, the solutions to the quadratic equation are $ x = 4 $ and $ x = 5 $. To verify these solutions, we can substitute them back into the original equation. For $ x = 4 $:

$ 4^2 - 9(4) + 20 = 0 $ simplifies to $ 16 - 36 + 20 = 0 $, confirming that $ 4 $ is indeed a solution.

Similarly, substituting $ x = 5 $ will also yield a true statement, confirming both values are correct. This process illustrates how factoring can effectively solve quadratic equations, allowing us to find the necessary values for $ x $ that satisfy the equation.

Problem

Solve the given quadratic equation by factoring. $3x^2+12x=0$

$x=3,x=4$

$x=0,x=-4$

$x=-3,x=-4$

$x=1,x=4$

Problem

Solve the given equation by factoring. $2x^2+7x+6=0$

$x=3,x=\frac67$

$x=-2,x=0$

$x=2,x=\frac32$

$x=-2,x=-\frac32$

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A quadratic equation is a polynomial equation of degree 2, typically written in the form $a^{2} x + b x + c = 0$ . The key difference from a linear equation, which is of degree 1 (e.g., $ax + b = 0$ ), is that a quadratic equation includes an $x^{2}$ term. This makes the graph of a quadratic equation a parabola, whereas the graph of a linear equation is a straight line.

To write a quadratic equation in standard form, you need to arrange it as $a^{2} x + b x + c = 0$ , where $a$ , $b$ , and $c$ are constants. Ensure all terms are on one side of the equation and ordered by descending power of $x$ . For example, $5 x^{2} - x + 3 = 0$ is in standard form.

In a quadratic equation of the form $a^{2} x + b x + c = 0$ , the coefficients are $a$ , $b$ , and $c$ . The coefficient $a$ is the number in front of $x^{2}$ , $b$ is the number in front of $x$ , and $c$ is the constant term. For example, in $3 x^{2} + 2 x - 6 = 0$ , $a$ is 3, $b$ is 2, and $c$ is -6.

To solve a quadratic equation by factoring, follow these steps: 1) Write the equation in standard form $a^{2} x + b x + c = 0$ . 2) Factor the quadratic expression. For example, $x^{2} + x - 6 = 0$ factors to $(x + 3) (x - 2) = 0$ . 3) Set each factor equal to zero: $x + 3 = 0$ and $x - 2 = 0$ . 4) Solve for $x$ : $x = - 3$ and $x = 2$ . These are the solutions.

A quadratic equation can have two solutions because it is a second-degree polynomial, meaning its graph is a parabola. When you solve the equation by factoring, you set each factor equal to zero, leading to two possible values for $x$ . For example, in the equation $x^{2} - 5 x + 6 = 0$ , factoring gives $(x - 2) (x - 3) = 0$ . Setting each factor to zero, you get $x = 2$ and $x = 3$ . These are the two solutions where the parabola intersects the x-axis.