xc = (y^2/x^2 -1)^1/2

how do you solve for y?

Question

xc = (y^2/x^2 -1)^1/2

how do you solve for y?

anonymous · Answer

You have to "cancel" the operation of square root - and then other actions. But first how does one neutralize the square root ? ....this is hint

anonymous · Answer

YOU RAISE TO THE SQUARE of course

adunb8 · Answer

oh i see ! !! square both sides lol

anonymous · Answer

This is tricky : IT USUSALLY "GIVES BIRTH" to unexistent additional solutions

adunb8 · Answer

???

anonymous · Answer

like x= 2 has only one TRUE solution, but if square both sides then
x^2 = 4 has two, one of which is artificial and does not exist in the original equation

adunb8 · Answer

i dont understand how my teacher got $$y = \pm x \sqrt{cx^2+1}$$

anonymous · Answer

He actually squared AND decided that BOTH solutions are legitimate.

adunb8 · Answer

can you show me the steps?

anonymous · Answer

After squaring  one obtainc y^2/x^2 - 1 = cx^2

anonymous · Answer

transfer 1 leftwise

anonymous · Answer

Multiply BOTH sdes by x^2

adunb8 · Answer

wouldnt it be x^2c^2?

anonymous · Answer

Yes it must

anonymous · Answer

the way u wrote it the solution of ur teacher is a wee-bit missing the truth

anonymous · Answer

Hell-oo ?

adunb8 · Answer

i still dont understand..

adunb8 · Answer

because when i did it i got like something way off

anonymous · Answer

$$c^2x^{2} +1 = \frac{ y^2 }{ x^2 }$$

anonymous · Answer

Now multpl. both sides by x^2

adunb8 · Answer

okay then that would be x^4c^2 + x^2

anonymous · Answer

One obtains
$$x^2(c^2x^2 +1) = y^2 $$

anonymous · Answer

YOU DO NOT OPEN THE BRACKETS

anonymous · Answer

Extract the ROOT ! (like "RELEASE THE CRACKEN !")

adunb8 · Answer

oh i see thank you! so you do sqrt to all of them then |dw:1346828060843:dw|