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7.7: Zero Product Property for Quadratic Equations

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The area of a particular rectangle was found to be A(w)=w^2-8w-58 . Determine the dimensions of the rectangle if the area was known to be 7 units.

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Khan Academy Factoring Special Products

Guidance

Recall that when solving an equation, you are trying to determine the values of the variable that make the equation true. For the equation 2x^2+10x+8=0 , x=-1 and x=-4 are both solutions. You can check this:

  • 2(-1)^2+10(-1)+8=2(1)-10+8=0
  • 2(-4)^2+10(-4)+8=2(16)-40+8=32-40+8=0

Here you will focus on solving quadratic equations. One of the methods for quadratic equations utilizes your factoring skills and a property called the zero product property .

If a\cdot b=0 , what can you say about a or b ? What you should realize is that either a or b have to be equal to 0, because that is the only way that their product will be 0. If both a and b were non-zero, then their product would have to be non-zero. This is the idea of the zero product property. The zero product property states that if the product of two quantities is zero, then one or both of the quantities must be zero.

The zero product property has to do with products being equal to zero. When you factor, you turn a quadratic expression into a product. If you have a quadratic expression equal to zero, you can factor it and then use the zero product property to solve. So, if you were given the equation 2x^2+5x-3=0 , first you would want to turn the quadratic expression into a product by factoring it:

2x^2+5x-3=(x+3)(2x-1)

You can rewrite the equation you are trying to solve as (x+3)(2x-1)=0 .

Now, you have the product of two binomials equal to zero. This means at least one of those binomials must be equal to zero. So, you have two mini-equations that you can solve to find the values of x that cause each binomial to be equal to zero.

  • x+3=0 , which means x=-3 OR
  • 2x-1=0 , which means x=\frac{1}{2}

The two solutions to the equation 2x^2+5x-3=0 are x=-3 and x=\frac{1}{2} .

Keep in mind that you can only use the zero product property if your equation is set equal to zero! If you have an equation not set equal to zero, first rewrite it so that it is set equal to zero. Then factor and use the zero product property.

Example A

Solve for x : x^2+5x+6=0 .

Solution: First, change x^2+5x+6 into a product so that you can use the zero product property. Change the expression into a product by factoring:

x^2+5x+6=(x+3)(x+2)

Next, rewrite the equation you are trying to solve:

x^2+5x+6=0 becomes (x+3)(x+2)=0 .

Finally, set up two mini-equations to solve in order to find the values of x that cause each binomial to be equal to zero.

  • x+3=0 , which means that x=-3
  • x+2=0 , which means that x=-2

The solutions are x=-3 or x=-2 .

Example B

Solve for x : 6x^2+x-15=0 .

In order to solve for x you need to factor the polynomial.

Solution: First, change 6x^2+x-15 into a product so that you can use the zero product property. Change the expression into a product by factoring:

6x^2+x-15=(3x+5)(2x-3)

Next, rewrite the equation you are trying to solve:

6x^2+x-15=0 becomes (3x+5)(2x-3)=0 .

Finally, set up two mini-equations to solve in order to find the values of x that cause each binomial to be equal to zero.

  • 3x+5=0 , which means that x=-\frac{5}{3}
  • 2x-3=0 , which means that x=\frac{3}{2}

The solutions are x=-\frac{5}{3} or x=\frac{3}{2} .

Example C

Solve for x : x^2+2x-35=0 .

Solution: First, change x^2+2x-35 into a product so that you can use the zero product property. Change the expression into a product by factoring:

x^2+2x-35=(x+7)(x-5)

Next, rewrite the equation you are trying to solve:

x^2+2x-35=0 becomes (x+7)(x-5)=0 .

Finally, set up two mini-equations to solve in order to find the values of x that cause each binomial to be equal to zero.

  • x+7=0 , which means that x=-7
  • x-5=0 , which means that x=5

The solutions are x=-7 or x=5 .

Concept Problem Revisited

The area of a particular rectangle was found to be A(w)=w^2-8w-58 . Determine the dimensions of the rectangle if the area was known to be 7 units.

In other words, you are being asked to solve the problem:

& w^2-8w-58=7\\& OR\\& w^2-8w-65=0

You can solve this problem by factoring and using the zero product property.

w^2-8w-65=0 becomes (w+5)(w-13)=0

& \quad (w+5)(w-13)=0\\&\qquad \swarrow \qquad \quad \searrow\\& w + 5 = 0 \qquad \quad w - 13 = 0\\&\quad \ \ w = -5 \quad or \qquad \ \ w = 13

Since you are asked for dimensions, a width of –5 units does not make sense. Therefore for the rectangle, the width would be 13 units.

Vocabulary

Quadratic Equation
A quadratic equation is an equation in which the highest power of a variable is 2. Standard form for a quadratic equation is ax^2+bx+c=0 .
Zero-product property
The zero-product property states that if two factors are multiplied together and their product is zero, then one of the factors must equal zero

Guided Practice

1. Solve for the variable in the polynomial: x^2+4x-21=0

2. Solve for the variable in the polynomial: 20m^2+11m-4=0

3. Solve for the variable in the polynomial: 2e^2+7e+6=0

Answers:

1. x^2+4x-21=(x-3)(x+7)

& \qquad (x-3)(x+7)=0\\&\quad \qquad \swarrow \qquad \quad \searrow\\&(x - 3) = 0 \qquad  (x+7) = 0\\&\qquad \ x = 3 \quad  \qquad \quad \ \ x = -7

2. 20m^2+11m-4=(4m-1)(5m+4)

& \qquad (4m-1)(5m+4)=0\\&\quad \qquad \swarrow \qquad \quad \searrow\\&4m - 1 = 0 \qquad  5m+4 = 0\\&\quad \ \ 4m = 1 \quad  \qquad \ \ 5m = -4\\&\qquad  m = \frac{1}{4} \quad  \qquad \quad  m = \frac{-4}{5}

3. 2e^2+7e+6=(2e+3)(e+2)

& \qquad (2e+3)(e+2)=0\\&\quad \qquad \swarrow \qquad \quad \searrow\\   & 2e+3 = 0 \qquad  \quad e+2 = 0\\&\quad \ \ 2e = -3 \quad  \qquad \quad e = -2\\&\qquad  e = \frac{-3}{2}

Practice

Solve for the variable in each of the following equations.

  1. (x+1)(x-3)=0
  2. (a+3)(a+5)=0
  3. (x-5)(x+4)=0
  4. (2t-4)(t+3)=0
  5. (x-8)(3x-7)=0
  6. x^2+x-12=0
  7. b^2+2b-24=0
  8. t^2+3t-18=0
  9. w^2+3w-108=0
  10. e^2-2e-99=0
  11. 6x^2-x-2=0
  12. 2d^2+14d-16=0
  13. 3s^2+20s+12=0
  14. 18x^2+12x+2=0
  15. 3j^2-17j+10=0

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Date Created:

Apr 30, 2013

Last Modified:

Jul 15, 2014
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