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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:
\[4x^2 - 3 = 64\]
shaun786
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4x^2 = 67
x^2 = 67/4
x = 67/4 , -67/4
need_help!
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Thanks =)
need_help!
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But 64 does not become 8?
Kreshnik
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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:
\[\LARGE \sqrt{(4x^2-3)^2}=64 \]
shaun786
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oh yeah forgot abt the squarerott. Thanks kreshnik
Kreshnik
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:)
need_help!
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Yes! That's the given.
need_help!
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Actually I don't know whether the square root is for everything or only for (4x^2-3)^2
need_help!
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But it does matter, right?
need_help!
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I mean if it was for everything then 64 will become 8?
lalaly
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What do u mean for everything?
need_help!
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for the whole equation
lalaly
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the whole sqrt is squared?
\[(\sqrt{4x^2-3})^2=64\]
need_help!
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Okay, sorry. I'm a bit confused. I just realised the question does not even have a square root, lol. Sorry!
lalaly
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haha so what is the question?
need_help!
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I have just edited the question.
eliassaab
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\[
(4x^2-3)^2=64 \\
(4x^2-3)=8 \\
4 x^2 = 11\\
x^2 = \frac {11}4\\
x=\pm \sqrt{\frac {11}4}\\
x =\pm \frac {\sqrt {11}} 2
\]
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Thank you so much =)