Question

Someone hel me solve this PLEASE!!

Answer 1

@ganeshie8 can you help plese ?

Answer 2

I would choose,
\[p=x^4\]

Answer 3

i dont understand

Answer 4

@DebbieG

Answer 5

The problem says you have to solve after choose an appropiate substitution. One appropiate substitution would be
\[p=x^4\]
You must plug this substitution in your original equation,
\[\sqrt{p}-2\sqrt[4]{p}-3=0\Rightarrow \sqrt{t^4}-2\sqrt[4]{t^4}-3=0\]

Answer 6

Sorry I put,
\[p=t^4\]

Answer 7

Once you solve it for t, then you must find p.

Answer 8

Do you understand it?

Answer 9

im understanding it better

Answer 10

Do you know how to solve it from this point?

Answer 11

so i got u=3 and u=-1 so i plug tht into original equatin ?

Answer 12

what do i do after

Answer 13

This is called an "equation of quadratic type". so with a proper substitution, you can solve as a quadratic equation.

It's as @John_ES says, although I would view the substitution a bit differently.

First rewrite the equation:
\(\Large p^{1/2}-2p^{1/4}-3=0\)

Now let \(\Large u=p^{1/4}\), then \(\Large u^2=(p^{1/4})^2=p^{1/2}\)

Makes those substitutions in your equation, and what do you have?

Answer 14

Now back-substitute your solutions for u, to find your solutions for p.

Answer 15

is it 81 and 1 ?

Answer 16

Yes, now you should put in the relation,
\[p=u^4\]
And obtain two values for p.

Answer 17

did that

Answer 18

Yes, they should be the values.

Answer 19

And you can always check them in the original equation, to make sure. :)

Answer 20

(The beauty of algebra! :)

Answer 21

81 and 1 is wrong

Answer 22

Yes, check them, because only one is the solution ;).

Answer 23

Right :)

Answer 24

its 81 and thankx guys