For what values of (2x+7)(x-5)=-43+jx does the equation have exactly one real solution?
Please help, I don't understand...
A quadratic equation can have exactly one solution if it is a perfect square.
(2x + 7)(x - 5) = -43 + jx
2x2 - 10x + 7x - 35 = -43 + jx [multiply out]
2x2 - 3x - 35 = -43 + jx [simplify]
2x2 - 3x - jx - 35 + 43 = 0 [subtract -43 + jx from both sides]
2x2 - (3 + j)x + 8 = 0 [factor and simplify]
2[ x2 - ( (3 + j)/2 )x + 4 ] = 0 [factor out the 2]
x2 - ( (3 + j)/2 )x + 4 = 0 [divide both sides by 2]
Now: x2 - 4x + 4 = (x - 2)2 = 0 and x2 + 4x + 4 = (x + 2)2 = 0 will have one real solution each.
So: either (3 + j)/2 = 4 or (3 + j)/2 = -4
Solving: 3 + j = 8 or 3 + j = -8
j = 5 or j = -11