x^2 + 4x - 31 = a
x^2 + 4x - 31 - a = 0
to have one root:
4^2 - 4(1)(-31-a) = 0
16 + 124 + 4a = 0
4a = -140
a = - 35
if f(x) = -35
x^2 + 4x - 31 = -35
x^2 + 4x + 4 = 0
(x+2)^2 = 0
x+2 = 0
x = -2 ----> only one root
Let f(x) = x^2 + 4x - 31. For what value of a is there exactly one real value of x such that f(x) = a?
Thank you!
11 answers
thats wrong
No, It's right.
It's wrong.
It's right.
It's Wrong
U already got it be4, its -35
Its right
No its wrong. -2 equals x, a = -35
Its not -2 its -35. Its wrong
Stop cheating! But, -35 is right.