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

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The square root and 2th power exclude each other as they are both doing the opposite. You therefore get: \[4x^2 - 3 = 64\]
4x^2 = 67 x^2 = 67/4 x = 67/4 , -67/4
Thanks =)

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Other answers:

But 64 does not become 8?
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: \[\LARGE \sqrt{(4x^2-3)^2}=64 \]
oh yeah forgot abt the squarerott. Thanks kreshnik
Yes! That's the given.
Actually I don't know whether the square root is for everything or only for (4x^2-3)^2
But it does matter, right?
I mean if it was for everything then 64 will become 8?
What do u mean for everything?
for the whole equation
the whole sqrt is squared? \[(\sqrt{4x^2-3})^2=64\]
Okay, sorry. I'm a bit confused. I just realised the question does not even have a square root, lol. Sorry!
haha so what is the question?
I have just edited the question.
\[ (4x^2-3)^2=64 \\ (4x^2-3)=8 \\ 4 x^2 = 11\\ x^2 = \frac {11}4\\ x=\pm \sqrt{\frac {11}4}\\ x =\pm \frac {\sqrt {11}} 2 \]
Thank you so much =)

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