Solving quadratic equations | Lesson (article) | Khan Academy (2024)

• $x$xx is the variable, which represents a number whose value we don't know yet.
• The $^2$2squared is the power or exponent. An exponent of $2$22 means the variable is .
• $3$33 and $-5$−5minus, 5 are the coefficients, or constant multiples of $x^2$x2x, squared and $x$xx. $3x^2$3x23, x, squared is a single , as is $-5x$−5xminus, 5, x.
• $-2$−2minus, 2 is a constant term.

In this lesson, we'll learn to:

1. Solve quadratic equations in several different ways
2. Determine the number of solutions to a quadratic equation without solving

You can learn anything. Let's do this!

Quadratic equations without $x$xx-terms such as $2x^2=32$2x2=322, x, squared, equals, 32 can be solved without setting a quadratic expression equal to $0$00. Instead, we can isolate $x^2$x2x, squared and use the square root operation to solve for $x$xx.

When solving quadratic equations by taking square roots, both the positive and negative square roots are solutions to the equation. This is because when we square a solution, the result is always positive.

For example, for the equation $x^2=4$x2=4x, squared, equals, 4, both $2$22 and $-2$−2minus, 2 are solutions:

• $2^2\stackrel{\checkmark}=4$22=✓42, squared, equals, start superscript, \checkmark, end superscript, 4
• $(-2)^2\stackrel{\checkmark}=4$(−2)2=✓4left parenthesis, minus, 2, right parenthesis, squared, equals, start superscript, \checkmark, end superscript, 4

When solving quadratic equations without $x$xx-terms:

1. Isolate $x^2$x2x, squared.
2. Take the square root of both sides of the equation. Both the positive and negative square roots are solutions.

Example: What values of $x$xx satisfy the equation $2x^2=18$2x2=182, x, squared, equals, 18 ?

The zero product property states that if $ab=0$ab=0a, b, equals, 0, then either $a$aa or $b$bb is equal to $0$00.

To solve a factored quadratic equation using the zero product property:

1. Set each factor equal to $0$00.
2. Solve the equations from Step 1. The solutions to the linear equations are also solutions to the quadratic equation.

Example: What are the solutions to the equation $(x-4)(3x+1)=0$(x−4)(3x+1)=0left parenthesis, x, minus, 4, right parenthesis, left parenthesis, 3, x, plus, 1, right parenthesis, equals, 0 ?

If we can write a quadratic expression as the product of two linear expressions (factors), then we can use those linear expressions to calculate the solutions to the quadratic equation.

In this lesson, we'll focus on factorable quadratic equations with $1$11 as the coefficient of the $x^2$x2x, squared term, such as $x^2-2x-3=0$x2−2x−3=0x, squared, minus, 2, x, minus, 3, equals, 0. For more advanced factoring techniques, including special factoring and factoring quadratic expressions with $x^2$x2x, squared coefficients other than $1$11, check out the Structure in expressions lesson.

Recognizing factors of quadratic expressions takes practice. The factors will be in the form $(x+a)(x+b)$(x+a)(x+b)left parenthesis, x, plus, a, right parenthesis, left parenthesis, x, plus, b, right parenthesis, where $a$aa and $b$bb fulfill the following criteria:

• The sum of $a$aa and $b$bb is equal to the coefficient of the $x$xx-term in the unfactored quadratic expression.
• The product of $a$aa and $b$bb is equal to the constant term of the unfactored quadratic expression.

For example, we can solve the equation $x^2-2x-3=0$x2−2x−3=0x, squared, minus, 2, x, minus, 3, equals, 0 by factoring $x^2-2x-3$x2−2x−3x, squared, minus, 2, x, minus, 3 into $(x+a)(x+b)$(x+a)(x+b)left parenthesis, x, plus, a, right parenthesis, left parenthesis, x, plus, b, right parenthesis, where:

• $a+b$a+ba, plus, b is equal to the coefficient of the $x$xx-term, $-2$−2minus, 2.
• $ab$aba, b is equal to the constant term, $-3$−3minus, 3.

$-3$−3minus, 3 and $1$11 would work:

• $-3+1=-2$−3+1=−2minus, 3, plus, 1, equals, minus, 2
• $(-3)(1)=-3$(−3)(1)=−3left parenthesis, minus, 3, right parenthesis, left parenthesis, 1, right parenthesis, equals, minus, 3

When solving factorable quadratic equations in the form $x^2+bx+c=0$x2+bx+c=0x, squared, plus, b, x, plus, c, equals, 0:

1. Rewrite the quadratic expression as the product of two factors. The two factors are linear expressions with an $x$xx-term and a constant term. The sum of the constant terms is equal to $b$bb, and the product of the constant terms is equal to $c$cc.
2. Set each factor equal to $0$00.
3. Solve the equations from Step 2. The solutions to the linear equations are also solutions to the quadratic equation.

Example: What are the solutions to the equation $x^2+4x-5=0$x2+4x−5=0x, squared, plus, 4, x, minus, 5, equals, 0 ?

Not all quadratic expressions are factorable, and not all factorable quadratic expressions are easy to factor. The quadratic formula gives us a way to solve any quadratic equation as long as we can plug the correct values into the formula and evaluate.

For $\purpleD{a}x^2+\maroonD{b}x+\tealE{c}=0$ax2+bx+c=0start color #7854ab, a, end color #7854ab, x, squared, plus, start color #ca337c, b, end color #ca337c, x, plus, start color #208170, c, end color #208170, equals, 0:

Note: the quadratic formula is not provided in the reference section of the SAT! You'll have to memorize the formula to use it.

To solve a quadratic equation using the quadratic formula:

1. Rewrite the equation in the form $ax^2+bx+c=0$ax2+bx+c=0a, x, squared, plus, b, x, plus, c, equals, 0.
2. Substitute the values of $a$aa, $b$bb, and $c$cc into the quadratic formula, shown below.

3. Evaluate $x$xx.

Example: What are the solutions to the equation $x^2-6x=9$x2−6x=9x, squared, minus, 6, x, equals, 9 ?

The $b^2-4ac$b2−4acb, squared, minus, 4, a, c portion of the quadratic formula is called the discriminant. The value of $b-4ac$b−4acb, minus, 4, a, c tells us the number of unique real solutions the equation has:

• If $b^2-4ac>0$b2−4ac>0b, squared, minus, 4, a, c, is greater than, 0, then $\sqrt{b^2-4ac}$b2−4ac​square root of, b, squared, minus, 4, a, c, end square root is a real number, and the quadratic equation has two real solutions, $\dfrac{-b-\sqrt{b^2-4ac}}{2a}$2a−b−b2−4ac​​start fraction, minus, b, minus, square root of, b, squared, minus, 4, a, c, end square root, divided by, 2, a, end fraction and $\dfrac{-b+\sqrt{b^2-4ac}}{2a}$2a−b+b2−4ac​​start fraction, minus, b, plus, square root of, b, squared, minus, 4, a, c, end square root, divided by, 2, a, end fraction.
• If $b^2-4ac=0$b2−4ac=0b, squared, minus, 4, a, c, equals, 0, then $\sqrt{b^2-4ac}$b2−4ac​square root of, b, squared, minus, 4, a, c, end square root is also $0$00, and the quadratic formula simplifies to $\dfrac{-b}{2a}$2a−b​start fraction, minus, b, divided by, 2, a, end fraction, which means the quadratic equation has one real solution.
• If $b^2-4ac<0$b2−4ac<0b, squared, minus, 4, a, c, is less than, 0, then $\sqrt{b^2-4ac}$b2−4ac​square root of, b, squared, minus, 4, a, c, end square root is an imaginary number, which means the quadratic equation has no real solutions.

For $ax^2+bx+c=0$ax2+bx+c=0a, x, squared, plus, b, x, plus, c, equals, 0:

• If $b^2-4ac>0$b2−4ac>0b, squared, minus, 4, a, c, is greater than, 0, then the equation has $2$22 unique real solutions.
• If $b^2-4ac=0$b2−4ac=0b, squared, minus, 4, a, c, equals, 0, then the equation has $1$11 unique real solution.
• If $b^2-4ac<0$b2−4ac<0b, squared, minus, 4, a, c, is less than, 0, then the equation has no real solution.