+0  
 
0
42
1
avatar+23 

For what values of $j$ does the equation $(2x+7)(x-5) = -43 + jx have exactly one real solution? If you find more than one, then list the values separated by commas.

 Jan 26, 2019
 #1
avatar+96208 
+1

(2x + 7) ( x - 5)  =   -43 + jx      simplify

 

2x^2 + 7x - 10x - 35 = -43 + jx

 

2x^2  - 3x - 35  =  -43 + jx

 

2x^2 - (3 + j)x + 8 = 0

 

This will have one real root when the discriminant = 0

 

So

 

(3 + j)^2 -   (4)(2)(8)  =  0

 

(3 + j)^2  =  64       take both roots

 

3 + j  =  8               or           3 + j = - 8

 

j = 5                      or           j = -11

 

 

cool cool cool

 Jan 26, 2019

46 Online Users

avatar
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.