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w What is twice the nonnegative difference between the solutions of the equation $x^2-7x-4=40$?

Guest Dec 29, 2017

Best Answer 

 #1
avatar+6955 
+1

First we need to find the solutions of the equation...

 

x2 - 7x - 4  =  40

                                    Subtract  40  from both sides so that the right side is zero.

x2 - 7x - 44  =  0

                                    Now we can factor the left side like this..

(x + 4)(x - 11)  =  0

                                    Set each factor equal to zero and solve each for  x .

 

x + 4  =  0         or          x - 11  =  0

x  =  -4              or          x  =  11

 

The solutions of the equation are  -4  and  11 .

 

Twice the nonnegative difference of  -4  and  11  is...

 

2 * | -4 - 11 |   =   2 * | -15 |   =   2 * 15   =   30

hectictar  Dec 29, 2017
edited by hectictar  Dec 29, 2017
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1+0 Answers

 #1
avatar+6955 
+1
Best Answer

First we need to find the solutions of the equation...

 

x2 - 7x - 4  =  40

                                    Subtract  40  from both sides so that the right side is zero.

x2 - 7x - 44  =  0

                                    Now we can factor the left side like this..

(x + 4)(x - 11)  =  0

                                    Set each factor equal to zero and solve each for  x .

 

x + 4  =  0         or          x - 11  =  0

x  =  -4              or          x  =  11

 

The solutions of the equation are  -4  and  11 .

 

Twice the nonnegative difference of  -4  and  11  is...

 

2 * | -4 - 11 |   =   2 * | -15 |   =   2 * 15   =   30

hectictar  Dec 29, 2017
edited by hectictar  Dec 29, 2017

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