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Maths Question

Given that y is negative , make y the subject of
y = √ay^2 + x

My answer was y= √-x/a-1 but it's wrong.
(edited 5 years ago)
1. you can't make anything the subject of √ay^2 + x because it's an expression not an equation.
2. your format is unclear since we can't tell what's square rooted in the original expression.
3. why are you asking us to do your homework for you.
Original post by nells45581
1. you can't make anything the subject of √ay^2 + x because it's an expression not an equation.
2. your format is unclear since we can't tell what's square rooted in the original expression.
3. why are you asking us to do your homework for you.


He is correct its an expression not an equation.
Original post by BrandonS03
He is correct its an expression not an equation.


*she :colondollar:
Original post by BrandonS03
He is correct its an expression not an equation.

She*
Error 404 equal sign not found

EDIT: it was prolly a typo so that + was an = in reality

EDIT 2: in the end it turns out OP forgot the = sign
(edited 5 years ago)
I'm going to assume the equation was: square root of (ay^2) = x

Because = and + are on the same key so was just a typo possibly.

Then start by squaring both sides and go from there.
Reply 7
Mb just edited forgot to put the y= and its the square root to the whole of ay^2 + x
(edited 5 years ago)
Original post by YamsAreYams
Mb just edited forgot to put the y= and its the square root to the whole of ay^2 + x


In that case, your answer is pretty much fine (though it's much nicer to write x1a\dfrac{x}{1-a} rather than xa1\dfrac{-x}{a-1})

Anyway, you are told that yy is negative, and here you have it as positive.

When you square root something, remember that y2=(something)    y=±(something)y^2 = \text{(something)} \implies y = \pm \sqrt{\text{(something)}}. But the question is more specific by saying that y is negative, so it can only be - in the front.
Reply 9
@RDKGames Ah thanks.

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