anonymous
  • anonymous
Solve (4x^2-3)^2=64
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
The square root and 2th power exclude each other as they are both doing the opposite. You therefore get: \[4x^2 - 3 = 64\]
anonymous
  • anonymous
4x^2 = 67 x^2 = 67/4 x = 67/4 , -67/4
anonymous
  • anonymous
Thanks =)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.