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10.8: Quadratic Formula

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What if you had a quadratic equation like x^2 + 5x + 2 that you could not easily factor? How could you use its coeffient values to solve it? After completing this Concept, you'll be able to use the quadratic formula to solve equations like this one.

Watch This

CK-12 Foundation: 1008S The Quadratic Formula

For more examples of solving quadratic equations using the quadratic formula, see the Khan Academy video at

Guidance

The Quadratic Formula is probably the most used method for solving quadratic equations. For a quadratic equation in standard form, ax^2 + bx + c = 0 , the quadratic formula looks like this:

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

This formula is derived by solving a general quadratic equation using the method of completing the square that you learned in the previous section.

We start with a general quadratic equation: ax^2 + bx + c = 0

Subtract the constant term from both sides: ax^2 + bx = -c

Divide by the coefficient of the x^2 term:

x^2 + \frac{b}{a} x = - \frac{c}{a}

Rewrite:

x^2 + 2 \left (\frac{b}{2a} \right ) x = - \frac{c}{a}

Add the constant \ \left (\frac{b}{2a} \right )^2 to both sides:

x^2 + 2 \left (\frac{b}{2a} \right )x + \left (\frac{b}{2a} \right )^2 = - \frac{c}{a} + \frac{b^2}{4a^2}

Factor the perfect square trinomial:

\left (x + \frac{b}{2a} \right )^2 = - \frac{4ac}{4a^2} + \frac{b^2}{4a^2}

Simplify:

\left (x + \frac{b}{2a} \right )^2 = \frac{b^2 - 4ac}{4a^2}

Take the square root of both sides:

x + \frac{b}{2a} = \sqrt{\frac{b^2 - 4ac}{4a^2}} \ \text{and} \ x + \frac{b}{2a} = - \sqrt{\frac{b^2 - 4ac}{4a^2}}

Simplify:

x + \frac{b}{2a} = \frac{\sqrt{b^2 - 4ac}}{2a} \ \text{and} \ x + \frac{b}{2a} = - \frac{\sqrt{b^2 - 4ac}}{2a}

x = - \frac{b}{2a} + \frac{\sqrt{b^2 - 4ac}}{2a} \ \text{and} \ x = - \frac{b}{2a} - \frac{\sqrt{b^2 - 4ac}}{2a}

x = \frac{-b + \sqrt{b^2 - 4ac}}{2a} \ \text{and} \ x = \frac{-b - \sqrt{b^2 - 4ac}}{2a}

This can be written more compactly as x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} .

You can see that the familiar formula comes directly from applying the method of completing the square. Applying the method of completing the square to solve quadratic equations can be tedious, so the quadratic formula is a more straightforward way of finding the solutions.

Solve Quadratic Equations Using the Quadratic Formula

To use the quadratic formula, just plug in the values of a, b, and c .

Example A

Solve the following quadratic equations using the quadratic formula.

a) 2x^2 + 3x + 1 = 0

b) x^2 - 6x + 5 = 0

c) -4x^2 + x + 1 = 0

Solution

Start with the quadratic formula and plug in the values of a, b and c .

a)  \text{Quadratic formula:} && x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\\text{Plug in the values} \ a = 2, \ b = 3, \ c = 1 && x &= \frac{-3 \pm \sqrt{(3)^2 - 4(2)(1)}}{2(2)}\\\text{Simplify:} && x &= \frac{-3 \pm \sqrt{9-8}}{4} = \frac{-3 \pm \sqrt{1}}{4}\\\text{Separate the two options:} && x &= \frac{-3 + 1}{4} \ \ \text{and} \ \ x = \frac{-3 - 1}{4}\\\text{Solve:} && x &= \frac{-2}{4} = - \frac{1}{2} \ \text{and} \ x = \frac{-4}{4} = -1

Answer: x = - \frac{1}{2} and x = -1

b) \text{Quadratic formula:} && x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\\text{Plug in the values} \ a = 1, \ b = -6, \ c = 5 && x &= \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(5)}}{2(1)}\\	 \text{Simplify:} && x &=\frac{6 \pm \sqrt{36 - 20}}{2} = \frac{6 \pm \sqrt{16}}{2}\\\text{Separate the two options:} && x &= \frac{6 + 4}{2} \ \text{and} \ x = \frac{6 -4}{2}\\\text{Solve:} && x &= \frac{10}{2} = 5 \ \text{and} \ x = \frac{2}{2} = 1

Answer: x = 5 and x = 1

c) \text{Quadratic formula:} && x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\\text{Plug in the values} \ a = -4, \ b = 1, \ c = 1 && x &= \frac{-1 \pm \sqrt{(1)^2 - 4(-4)(1)}}{2(-4)}\\\text{Simplify:} && x &= \frac{-1 \pm \sqrt{1 + 16}}{-8} = \frac{-1 \pm \sqrt{17}}{-8}\\\text{Separate the two options:} && x &= \frac{-1 + \sqrt{17}}{-8} \ \text{and} \ x = \frac{-1 - \sqrt{17}}{-8}\\\text{Solve:} && x &= -.39 \ \text{and} \ x = .64

Answer: x = -.39 and x = .64

Often when we plug the values of the coefficients into the quadratic formula, we end up with a negative number inside the square root. Since the square root of a negative number does not give real answers, we say that the equation has no real solutions. In more advanced math classes, you’ll learn how to work with “complex” (or “imaginary”) solutions to quadratic equations.

Example B

Use the quadratic formula to solve the equation x^2 + 2x + 7 = 0 .

Solution

\text{Quadratic formula:} && x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\\text{Plug in the values} \ a = 1, \ b = 2, \ c = 7 && x &= \frac{-2 \pm \sqrt{(2)^2 - 4(1)(7)}}{2(1)}\\\text{Simplify:} && x &= \frac{-2 \pm \sqrt{4 - 28}}{2} = \frac{-2 \pm \sqrt{-24}}{2}

Answer: There are no real solutions.

To apply the quadratic formula, we must make sure that the equation is written in standard form. For some problems, that means we have to start by rewriting the equation.

Finding the Vertex of a Parabola with the Quadratic Formula

Sometimes a formula gives you even more information than you were looking for. For example, the quadratic formula also gives us an easy way to locate the vertex of a parabola.

Remember that the quadratic formula tells us the roots or solutions of the equation ax^2 + bx + c = 0 . Those roots are  x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, , and we can rewrite that as x = \frac{-b}{2a} \pm \frac{\sqrt{b^2 - 4ac}} {2a}.

Recall that the roots are symmetric about the vertex. In the form above, we can see that the roots of a quadratic equation are symmetric around the x- coordinate \frac{-b}{2a} , because they are \frac{\sqrt{b^2 - 4ac}} {2a} units to the left and right (recall the \pm sign) from the vertical line x = \frac{-b}{2a} .

Example C

In the equation x^2 - 2x - 3 = 0 , the roots -1 and 3 are both 2 units from the vertical line x = 1 , as you can see in the graph below:

Watch this video for help with the Examples above.

CK-12 Foundation: 1008 The Quadratic Formula

Vocabulary

  • For a quadratic equation in standard form, ax^2 + bx + c = 0 , the quadratic formula looks like this:

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

  • The quadratic formula tells us the roots or solutions of the equation ax^2 + bx + c = 0 . Those roots are  x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} , and we can rewrite that as x = \frac{-b}{2a} \pm \frac{\sqrt{b^2 - 4ac}} {2a}.
  • The roots are symmetric about the vertex . In the form above, we can see that the roots of a quadratic equation are symmetric around the x- coordinate \frac{-b}{2a} , because they are \frac{\sqrt{b^2 - 4ac}} {2a} units to the left and right (recall the \pm sign) from the vertical line x = \frac{-b}{2a} .

Guided Practice

Solve the following equations using the quadratic formula.

a) x^2 - 6x = 10

b) -8x^2 = 5x + 6

Solution

a) \text{Re-write the equation in standard form:} && x^2 - 6x - 10 &= 0\\\text{Quadratic formula:} && x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\\text{Plug in the values} \ a = 1, \ b = -6, \ c = -10 && x &= \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-10)}}{2(1)}\\\text{Simplify:} && x &= \frac{6 \pm \sqrt{36 + 40}}{2} = \frac{6 \pm \sqrt{76}}{2}\\\text{Separate the two options:} && x &= \frac{6 + \sqrt{76}}{2} \ \text{and} \ x = \frac{6 - \sqrt{76}}{2}\\\text{Solve:} && x &= 7.36 \ \text{and} \ x = -1.36

Answer: x = 7.36 and x = -1.36

b) \text{Re-write the equation in standard form:} && 8x^2+5x+6 &= 0\\\text{Quadratic formula:} && x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\\text{Plug in the values} \ a = 8, \ b = 5, \ c = 6 && x &= \frac{-5 \pm \sqrt{(5)^2 - 4(8)(6)}}{2(8)}\\\text{Simplify:} && x &= \frac{-5 \pm \sqrt{25 - 192}}{16} = \frac{-5 \pm \sqrt{-167}}{16}

Answer: no real solutions

Practice

Solve the following quadratic equations using the quadratic formula.

  1. x^2 + 4x - 21 = 0
  2. x^2 - 6x = 12
  3. 3x^2 - \frac{1}{2}x = \frac{3}{8}
  4. 2x^2 + x - 3 = 0
  5. -x^2 - 7x + 12 = 0
  6. -3x^2 + 5x = 2
  7. 4x^2 = x
  8. x^2 + 2x + 6 = 0
  9. 5x^2 -2x + 100 = 0
  10. 100x^2 +10x + 70 = 0

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Date Created:

Oct 01, 2012

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Sep 15, 2014
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