A quadratic equation is polynomial equation with a degree of two, meaning the highest valued exponent in the equation is a two. Which of the following are quadratic equations?
A) \( 4 = 3x - 2 \)
B) \( 8 = -4x^2 + 6x - 7 \)
C) \( 1 = 9x^2 \)
D) \( 7x - 6 = 15 \)
E) \( 0= -|x - 6| + 3 \)
The equations from B and C are quadratic equations. Each equation has a term in which the highest degree is a \(2\). In this target, you are learning how to solve a quadratic equation. Why would we want to know how to do this? Solving a quadratic equation reveals the "zeros" of the function. This is important for graphing a quadratic function which we will be doing in the next unit Graphing Quadratic Functions There are several methods you can use to solve a quadratic equation: factoring, using square roots, completing the square, and using the quadratic formula. Below are steps and examples on how to solve a quadratic equation for each method. To learn when to use each method, see Solving Quadratic Equations Target B Guided Learning.
Method 1: Factoring
1. Make sure one side of the equation is equal to zero.
2. Check to see if you can factor out a GCF (if you factor out a GCF and there is no longer an \( x^2 \)-term, quadratic term,
proceed to step 4).
3. Factor – Difference of Two Squares, Perfect Square Trinomial, or Guess and Check.
4. Set each factor equal to zero by using the Zero Product Property.
5. Solve each equation.
A) \( 4 = 3x - 2 \)
B) \( 8 = -4x^2 + 6x - 7 \)
C) \( 1 = 9x^2 \)
D) \( 7x - 6 = 15 \)
E) \( 0= -|x - 6| + 3 \)
The equations from B and C are quadratic equations. Each equation has a term in which the highest degree is a \(2\). In this target, you are learning how to solve a quadratic equation. Why would we want to know how to do this? Solving a quadratic equation reveals the "zeros" of the function. This is important for graphing a quadratic function which we will be doing in the next unit Graphing Quadratic Functions There are several methods you can use to solve a quadratic equation: factoring, using square roots, completing the square, and using the quadratic formula. Below are steps and examples on how to solve a quadratic equation for each method. To learn when to use each method, see Solving Quadratic Equations Target B Guided Learning.
Method 1: Factoring
1. Make sure one side of the equation is equal to zero.
2. Check to see if you can factor out a GCF (if you factor out a GCF and there is no longer an \( x^2 \)-term, quadratic term,
proceed to step 4).
3. Factor – Difference of Two Squares, Perfect Square Trinomial, or Guess and Check.
4. Set each factor equal to zero by using the Zero Product Property.
5. Solve each equation.
Solving Quadratic Equations by Factoring Quick Check
1) \( -8x^2 + 20x = 0 \)
2) \( 2x^2 - 21x + 25 = -15 \)
Solving Quadratic Equations by Factoring Quick Check Solutions
1) \( -8x^2 + 20x = 0 \)
2) \( 2x^2 - 21x + 25 = -15 \)
Solving Quadratic Equations by Factoring Quick Check Solutions
Method 2: Using Square Roots
1. Isolate the “squared” expression.
2. Take the square root of both sides of the equation.
3. Solve for the variable.
1. Isolate the “squared” expression.
2. Take the square root of both sides of the equation.
3. Solve for the variable.
Solving Quadratic Equations Using Square Roots Quick Check
1) \( 6x^2 - 15 = 27 \)
2) \( 4(x - 2)^2 = 20 \)
Solving Quadratic Equations Using Square Roots Quick Check Solutions
1) \( 6x^2 - 15 = 27 \)
2) \( 4(x - 2)^2 = 20 \)
Solving Quadratic Equations Using Square Roots Quick Check Solutions
Method 3: Completing the Square
1. Isolate the \( x^2 \)-term and the x-term on one side of the equation (isolate the constant on the other side).
2. Make sure leading coefficient is \(1\).
3. Complete the square: divide the coefficient of the x-term (linear term) by \(2\), square this number, and add this
number to both sides of the equation to write an equivalent equation.
4. Write the trinomial as a perfect square: \( (x \space )^2 \).
5. Solve the equation using Square Roots.
1. Isolate the \( x^2 \)-term and the x-term on one side of the equation (isolate the constant on the other side).
2. Make sure leading coefficient is \(1\).
3. Complete the square: divide the coefficient of the x-term (linear term) by \(2\), square this number, and add this
number to both sides of the equation to write an equivalent equation.
4. Write the trinomial as a perfect square: \( (x \space )^2 \).
5. Solve the equation using Square Roots.
Solving Quadratic Equations by Completing the Square Quick Check
1) \( x^2 - 14x + 3 = 15 \)
2) \( 2x^2 + 16x - 8 = 0 \)
Solving Quadratic Equations by Completing the Square Quick Check Solutions
1) \( x^2 - 14x + 3 = 15 \)
2) \( 2x^2 + 16x - 8 = 0 \)
Solving Quadratic Equations by Completing the Square Quick Check Solutions
Method 4: Quadratic Formula
The quadratic formula is another approach to solving a quadratic equation.
The quadratic formula is another approach to solving a quadratic equation.
The quadratic formula: If \( ax^2 + bx + c =0 \), then \( x= \large\frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
To help you remember the formula, you can sing the following words to the tune of "Pop Goes the Weasel":
x equals negative b
plus or minus the square root
of b squared minus four a c
all over two a
Listen to the song
plus or minus the square root
of b squared minus four a c
all over two a
Listen to the song
If you aren't up for the song and want to understand why the formula is what it is, here is the derivation.
\( \begin{align*}
&ax^{2}+bx+c=0 \\
&x^{2}+\frac{b}{a}x+\frac{c}{a}=0 &&\text{divide by a}\\
&x^{2}+\frac{b}{a}x=-\frac{c}{a} &&\text{subtract $\frac{c}{a}$ from both sides} \\
&x^{2}+\frac{b}{a}x+\left(\frac{b}{2a} \right)^{2}=-\frac{c}{a}+\left(\frac{b}{2a} \right)^{2} &&\text{complete the square}\\
& \left( x+\frac{b}{2a} \right)^{2}=-\frac{4ac}{4a^{2}}+\frac{b^{2}}{4a^{2}} &&\text{factor LHS, obtain common denominator on RHS}\\
& \left(x+\frac{b}{2a} \right)^{2}=\frac{b^{2}-4ac}{4a^{2}} &&\text{add fractions on RHS}\\
&x+\frac{b}{2a}=\pm \frac{\sqrt{b^{2}-4ac}}{2a} &&\text{take the square root of both sides}\\
&x=-\frac{b}{2a}\pm \frac{\sqrt{b^{2}-4ac}}{2a} &&\text{subtract $\frac{b}{2a}$ from both sides}\\
&x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a} &&\text{add fractions on RHS}
\end{align*} \)
&ax^{2}+bx+c=0 \\
&x^{2}+\frac{b}{a}x+\frac{c}{a}=0 &&\text{divide by a}\\
&x^{2}+\frac{b}{a}x=-\frac{c}{a} &&\text{subtract $\frac{c}{a}$ from both sides} \\
&x^{2}+\frac{b}{a}x+\left(\frac{b}{2a} \right)^{2}=-\frac{c}{a}+\left(\frac{b}{2a} \right)^{2} &&\text{complete the square}\\
& \left( x+\frac{b}{2a} \right)^{2}=-\frac{4ac}{4a^{2}}+\frac{b^{2}}{4a^{2}} &&\text{factor LHS, obtain common denominator on RHS}\\
& \left(x+\frac{b}{2a} \right)^{2}=\frac{b^{2}-4ac}{4a^{2}} &&\text{add fractions on RHS}\\
&x+\frac{b}{2a}=\pm \frac{\sqrt{b^{2}-4ac}}{2a} &&\text{take the square root of both sides}\\
&x=-\frac{b}{2a}\pm \frac{\sqrt{b^{2}-4ac}}{2a} &&\text{subtract $\frac{b}{2a}$ from both sides}\\
&x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a} &&\text{add fractions on RHS}
\end{align*} \)
Steps for solving a quadratic equation using the quadratic formula:
1. Make sure the equation is in standard form (descending order) and equal to zero.
2. Identify the \( a, b, \) and \( c \).
3. Substitute \( a, b, \) and \( c \) into the quadratic formula.
4. Evaluate to find the value of the variable.
1. Make sure the equation is in standard form (descending order) and equal to zero.
2. Identify the \( a, b, \) and \( c \).
3. Substitute \( a, b, \) and \( c \) into the quadratic formula.
4. Evaluate to find the value of the variable.
Quick Check
1) \( x^2 - 3x - 7 = 0 \)
2) \( 3x^2 -11x + 14 = 4 \)
Solving Quadratic Equations Using the Quadratic Formula Quick Check Solutions
1) \( x^2 - 3x - 7 = 0 \)
2) \( 3x^2 -11x + 14 = 4 \)
Solving Quadratic Equations Using the Quadratic Formula Quick Check Solutions