Question

Solve (4x^2-3)^2=64

Answer 1

The square root and 2th power exclude each other as they are both doing the opposite. You therefore get:

\[4x^2 - 3 = 64\]

Answer 2

4x^2 = 67
x^2 = 67/4
x = 67/4 , -67/4

Answer 3

Thanks =)

Answer 4

But 64 does not become 8?

Answer 5

shaun has it right untill he gets rid of power
it should be:

\[\LARGE x^2=\frac{67}{4}\]
then...
\[\LARGE x=\pm \sqrt{\frac{67}{4}}\]
but first of all... just want to ask you.
is this the given:
\[\LARGE \sqrt{(4x^2-3)^2}=64 \]

Answer 6

oh yeah forgot abt the squarerott. Thanks kreshnik

Answer 7

:)

Answer 8

Yes! That's the given.

Answer 9

Actually I don't know whether the square root is for everything or only for (4x^2-3)^2

Answer 10

But it does matter, right?

Answer 11

I mean if it was for everything then 64 will become 8?

Answer 12

What do u mean for everything?

Answer 13

for the whole equation

Answer 14

the whole sqrt is squared?
\[(\sqrt{4x^2-3})^2=64\]

Answer 15

Okay, sorry. I'm a bit confused. I just realised the question does not even have a square root, lol. Sorry!

Answer 16

haha so what is the question?

Answer 17

I have just edited the question.

Answer 18

\[
(4x^2-3)^2=64 \\
(4x^2-3)=8 \\
4 x^2 = 11\\
x^2 = \frac {11}4\\
x=\pm \sqrt{\frac {11}4}\\
x =\pm \frac {\sqrt {11}} 2

\]

Answer 19

Thank you so much =)