+0  
 
0
145
1
avatar

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+7154 
+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
 #1
avatar+7154 
+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

15 Online Users

avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.