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10.5: Solving Quadratic Equations by the Quadratic Formula

Created by: CK-12

Learning Objectives

  • Solve quadratic equations using the quadratic formula.
  • Identify and choose methods for solving quadratic equations.
  • Solve real-world problems using functions by completing the square.

Introduction

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

\text{Divide by the coefficient of the} \ x^2 \ \text{term:} && x^2 + \frac{b}{a} x &= - \frac{c}{a}\\\text{Rewrite:} && x^2 + 2 \left (\frac{b}{2a} \right ) x &= - \frac{c}{a}\\\text{Add the constant} \ \left (\frac{b}{2a} \right )^2 \ \text{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}\\\text{Factor the perfect square trinomial:} && \left (x + \frac{b}{2a} \right )^2 &= - \frac{4ac}{4a^2} + \frac{b^2}{4a^2}\\\text{Simplify:} && \left (x + \frac{b}{2a} \right )^2 &= \frac{b^2 - 4ac}{4a^2}\\\text{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}}\\\text{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 1

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 2

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.

Example 3

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

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

. This video isn’t necessarily different from the examples above, but it does help reinforce the procedure of using the quadratic formula to solve equations.

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}. For example, 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:

Identify and Choose Methods for Solving Quadratic Equations.

In mathematics, you’ll need to solve quadratic equations that describe application problems or that are part of more complicated problems. You’ve learned four ways of solving a quadratic equation:

  • Factoring
  • Taking the square root
  • Completing the square
  • Quadratic formula

Usually you’ll have to decide for yourself which method to use. However, here are some guidelines as to which methods are better in different situations.

Factoring is always best if the quadratic expression is easily factorable. It is always worthwhile to check if you can factor because this is the fastest method. Many expressions are not factorable so this method is not used very often in practice.

Taking the square root is best used when there is no x-term in the equation.

Completing the square can be used to solve any quadratic equation. This is usually not any better than using the quadratic formula (in terms of difficult computations), but it is very useful if you need to rewrite a quadratic function in vertex form. It’s also used to rewrite the equations of circles, ellipses and hyperbolas in standard form (something you’ll do in algebra II, trigonometry, physics, calculus, and beyond).

Quadratic formula is the method that is used most often for solving a quadratic equation. When solving directly by taking square root and factoring does not work, this is the method that most people prefer to use.

If you are using factoring or the quadratic formula, make sure that the equation is in standard form.

Example 4

Solve each quadratic equation.

a) x^2 - 4x - 5 = 0

b) x^2 = 8

c) -4x^2 + x = 2

d) 25x^2 - 9 = 0

e) 3x^2 = 8x

Solution

a) This expression if easily factorable so we can factor and apply the zero-product property:

& \text{Factor:} && (x - 5)(x + 1) = 0\\& \text{Apply zero-product property:} && x - 5 = 0 \quad \text{and} \quad x + 1 = 0\\& \text{Solve:} && x = 5 \qquad \ \ \text{and} \quad x = -1

Answer: x = 5 and x = -1

b) Since the expression is missing the x term we can take the square root:

Take the square root of both sides: x = \sqrt{8} and x = - \sqrt{8}

Answer: x = 2.83 and x = -2.83

c) Re-write the equation in standard form: -4x^2 + x - 2 = 0

It is not apparent right away if the expression is factorable so we will use the quadratic formula:

\text{Quadratic formula:} && x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\\text{Plug in the values} \ a = -4, \ b = 1, \ c = -2: && x &= \frac{-1 \pm \sqrt{1^2 - 4(-4)(-2)}}{2(-4)}\\\text{Simplify:} && x &= \frac{-1 \pm \sqrt{1 - 32}}{-8} = \frac{-1 \pm \sqrt{-31}}{-8}

Answer: no real solution

d) This problem can be solved easily either with factoring or taking the square root. Let’s take the square root in this case:

\text{Add} \ 9 \ \text{to both sides of the equation:} && 25x^2 &= 9\\\text{Divide both sides by} \ 25: && x^2 &= \frac{9}{25}\\\text{Take the square root of both sides:} && x &= \sqrt{\frac{9}{25}} \ \text{and} \ x=-\sqrt{\frac{9}{25}}\\\text{Simplify:} && x &= \frac{3}{5} \ \text{and} \ x=-\frac{3}{5}

Answer: x = \frac{3}{5} and x = -\frac{3}{5}

e) \text{Re-write the equation in standard form:} && 3x^2 - 8x &= 0\\\text{Factor out common} \ x \ \text{term:} && x(3x-8) &= 0\\\text{Set both terms to zero:} && x &= 0 \ \text{and} \ 3x = 8\\\text{Solve:} && x &= 0 \ \text{and} \ x = \frac{8}{3} = 2.67

Answer: x =  0 and x = 2.67

Solve Real-World Problems Using Quadratic Functions by any Method

Here are some application problems that arise from number relationships and geometry applications.

Example 5

The product of two positive consecutive integers is 156. Find the integers.

Solution

Define: Let x = the smaller integer

Then x + 1 = the next integer

Translate: The product of the two numbers is 156. We can write the equation:

x(x + 1) = 156

Solve:

x^2 + x & = 156\\x^2 + x - 156 & = 0

Apply the quadratic formula with: a = 1, \ b = 1, \ c = -156

x & = \frac{-1 \pm \sqrt{1^2 - 4(1)(-156)}}{2(1)}\\x & = \frac{-1 \pm \sqrt{625}}{2} = \frac{-1 \pm 25}{2} \\x & = \frac{-1 + 25}{2} \quad \text{and} \quad x = \frac{-1 - 25}{2}\\x & = \frac{24}{2} = 12 \quad \text{and} \quad x = \frac{-26}{2} = -13

Since we are looking for positive integers, we want x = 12. So the numbers are 12 and 13.

Check: 12 \times 13 = 156. The answer checks out.

Example 6

The length of a rectangular pool is 10 meters more than its width. The area of the pool is 875 square meters. Find the dimensions of the pool.

Solution

Draw a sketch:

Define: Let x = the width of the pool

Then x + 10 = the length of the pool

Translate: The area of a rectangle is A = \text{length} \times \text{width}, so we have x(x +10) = 875.

Solve:

x^2 + 10x &= 875\\		 x^2 + 10x - 875 &= 0

Apply the quadratic formula with a = 1, \ b = 10 and c = -875

x & = \frac{-10 \pm \sqrt{(10)^2 - 4(1)(-875)}}{2(1)}\\x & = \frac{-10 \pm \sqrt{100 + 3500}}{2}\\x & = \frac{-10 \pm \sqrt{3600}}{2} = \frac{-10 \pm 60}{2}\\x & = \frac{-10 + 60}{2} \ \text{and} \ x = \frac{-10 - 60}{2}\\x & = \frac{50}{2} = 25 \ \text{and} \ x = \frac{-70}{2} = -35

Since the dimensions of the pool should be positive, we want x = 25 \ meters. So the pool is 25 \ meters \times 35 \ meters.

Check: 25 \times 35 = 875 \ m^2. The answer checks out.

Example 7

Suzie wants to build a garden that has three separate rectangular sections. She wants to fence around the whole garden and between each section as shown. The plot is twice as long as it is wide and the total area is 200 ft^2. How much fencing does Suzie need?

Solution

Define: Let x = the width of the plot

Then 2x = the length of the plot

Translate: area of a rectangle is A = \text{length} \times \text{width}, so

x(2x) = 200

Solve: 2x^2 = 200

Solve by taking the square root:

x^2 & = 100\\x & = \sqrt{100} \ \text{and} \ x = - \sqrt{100}\\x & = 10 \ \text{and} \ x = - 10

We take x = 10 since only positive dimensions make sense.

The plot of land is 10 \ feet \times 20 \ feet.

To fence the garden the way Suzie wants, we need 2 lengths and 4 widths = 2(20) + 4(10) = 80 feet of fence.

Check: 10 \times 20 = 200 \ ft^2 and 2(20) + 4(10) = 80 \ feet. The answer checks out.

Example 8

An isosceles triangle is enclosed in a square so that its base coincides with one of the sides of the square and the tip of the triangle touches the opposite side of the square. If the area of the triangle is 20 \ in^2 what is the length of one side of the square?

Solution

Draw a sketch:

Define: Let x = base of the triangle

Then x = height of the triangle

Translate: Area of a triangle is  \frac{1}{2} \times base \times height, so  \frac{1}{2} \cdot x \cdot x = 20

Solve: \frac{1}{2}x^2 = 20

Solve by taking the square root:

x^2 & = 40\\x & = \sqrt{40} \ \text{and} \ x = - \sqrt{40}\\x & = 6.32 \ \text{and} \ x = -6.32

The side of the square is 6.32 inches. That means the area of the square is (6.32)^2 = 40 \ in^2, twice as big as the area of the triangle.

Check: It makes sense that the area of the square will be twice that of the triangle. If you look at the figure you can see that you could fit two triangles inside the square.

Review Questions

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

Solve the following quadratic equations using the method of your choice.

  1. x^2 - x = 6
  2. x^2 - 12 = 0
  3. -2x^2 + 5x - 3 = 0
  4. x^2 + 7x - 18 = 0
  5. 3x^2 + 6x = - 10
  6. -4x^2 + 4000x = 0
  7. -3x^2 + 12x + 1 = 0
  8. x^2 + 6x + 9 = 0
  9. 81x^2 + 1 = 0
  10. -4x^2 + 4x = 9
  11. 36x^2 - 21 = 0
  12. x^2 - 2x - 3 = 0
  13. The product of two consecutive integers is 72. Find the two numbers.
  14. The product of two consecutive odd integers is 1 less than 3 times their sum. Find the integers.
  15. The length of a rectangle exceeds its width by 3 inches. The area of the rectangle is 70 square inches, find its dimensions.
  16. Angel wants to cut off a square piece from the corner of a rectangular piece of plywood. The larger piece of wood is 4 \ feet \times 8 \ feet and the cut off part is \frac{1}{3} of the total area of the plywood sheet. What is the length of the side of the square?
  17. Mike wants to fence three sides of a rectangular patio that is adjacent the back of his house. The area of the patio is 192 \ ft^2 and the length is 4 feet longer than the width. Find how much fencing Mike will need.

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