need_help! Group Title Solve (4x^2-3)^2=64 2 years ago 2 years ago

1. Retract Group Title

The square root and 2th power exclude each other as they are both doing the opposite. You therefore get: $4x^2 - 3 = 64$

2. shaun786 Group Title

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

3. need_help! Group Title

Thanks =)

4. need_help! Group Title

But 64 does not become 8?

5. Kreshnik Group Title

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: $\LARGE \sqrt{(4x^2-3)^2}=64$

6. shaun786 Group Title

oh yeah forgot abt the squarerott. Thanks kreshnik

7. Kreshnik Group Title

:)

8. need_help! Group Title

Yes! That's the given.

9. need_help! Group Title

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

10. need_help! Group Title

But it does matter, right?

11. need_help! Group Title

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

12. lalaly Group Title

What do u mean for everything?

13. need_help! Group Title

for the whole equation

14. lalaly Group Title

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

15. need_help! Group Title

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

16. lalaly Group Title

haha so what is the question?

17. need_help! Group Title

I have just edited the question.

18. eliassaab Group Title

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

19. need_help! Group Title

Thank you so much =)