Asked by Candice

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!
Answered by Reiny
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
Answered by Anonymous
thats wrong
Answered by No Name
No, It's right.
Answered by Anonymous
It's wrong.
Answered by lel
It's right.
Answered by Someone
It's Wrong
Answered by Someone
U already got it be4, its -35
Answered by Anonymous
Its right
Answered by Memer
No its wrong. -2 equals x, a = -35
Answered by Taco
Its not -2 its -35. Its wrong
Answered by AoPS
Stop cheating! But, -35 is right.
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