Question

Can someone help me with this question? Find all values of a, for which the equation: square root of(a*x^2+ax+2)=ax+2 has a unique root.

Answer 1

a*x^2+ax+2=ax+2
ax^2 = 0
x^2 = 0
x = 0

no matter what the value of a is, the equation always has a unique solution. Namely 0

Answer 2

but there is a square root so I would get ax^2+ax+2=(ax+2)^2

Answer 3

didn't see the square root -.- pft.. i'm off

Answer 4

After some simplification, the first line below will transform to the second line
\[
(a x^2+a x+2)=(a x+2)^2 \\

(a - a^2) x^2-3 a x -2 =0

\]
One a=1 will make it a first degree equation with one root.
a=0 does not.
There is still another a. Can you find it?

Answer 5

The discriminant of the quadratic equation above is
\[
(-3 a)^2 -4(a-a^2)(-2)=8a +a^2= a(8+a)=0
a=0\\
a=-8

\]
a=0 is not good.
a =-8 will give you double root.

Answer 6

for a=-8 the quadratic becomes
\[
-2 + 24 x - 72 x^2 = 0\\
36 x^2 - 12 x +1=0\\
(6x-1)^2=0\\
x= \frac 1 6
\]