+321 For what values of j does the equation ( 2 x + 7 )( x - 5) = - 43 + jx have exactly one real solution?
+59 ( 2 x + 7 )( x - 5) = - 43 + jx
2x2-3x-35-jx=-43
2x2-3x-35-jx+43=-43+43
2x2-(j+3)x+8=0
+130071 (2x + 7) ( x - 5) = -43 + jx expand
2x^2 -3x - 35 = -43 + jx rearrange as
2x^2 - ( 3 + j)x + 8 = 0
This will have one real root if the discriminant = 0 ....so....
(3 + j)^2 - 4(2) ( 8) = 0
j^2 + 6j + 9 - 64 = 0
j^2 + 6j - 55 = 0 factor as
(j + 11) ( j - 5) = 0
Set both factors to 0 and solve for j
We will have one real root whenever j = -11 or when j = 5
