Algebra problem confused by teacher's working

Mattatcomputer

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Hi, I hope I've got the right place to post. I've a test tomorrow and on the mock this question came up and I am confused about it.

The question is 'Make y the subject of: 6t2 = 5ay2 '

My working so far is divide both sides by 5a (6t2/5a=y2). Then do the square root of both sides (not sure if I do square root of just 6t2 or the whole side (6t2/5a)? Giving me 2.44t/5a = y.

My teacher has done y2=6t2/5a > y=square root of (6t2/5a) = t square root of (6/5a) I am most confused about this t square root. So his answer is y=t square root of 6/5a

Appreciate any help here, cheers

 
The question is 'Make y the subject of: 6t2 = 5ay2 '

My working so far is divide both sides by 5a (6t2/5a=y2). Then do the square root of both sides (not sure if I do square root of just 6t2 or the whole side (6t2/5a)? Giving me 2.44t/5a = y.

My teacher has done y2=6t2/5a > y=square root of (6t2/5a) = t square root of (6/5a) I am most confused about this t square root. So his answer is y=t square root of 6/5a

First, whenever you "do the same thing to both sides", that always means "to the entire value of each side", not just "to part of each side". So you have to take the square root of ALL of 6t2/5a. And you can see that the teacher did that.

Next, you should have learned something about simplifying a radical. Recall that sqrt(ab) = sqrt(a) sqrt(b) -- that is, the root of a product is the product of the roots. When you have a factor that is a perfect square, like t2 here, you can break that out like this: sqrt(
6t2/5a) = sqrt(t2 * 6/5a) = sqrt(t2) sqrt(6/5a) = t sqrt(6/5a) ... as long as t is known to be positive. (If that is not known, then it would be better to leave it alone.)

I see that in your own work, you actually did that! You took the square root of 6, 2.44, and the square root of t2, t. The only thing wrong there is that in this kind of problem we want exact answers, so we avoid decimal approximations. (You'll be told when to give a decimal answer, usually in applied problems.)
 
Hi, I hope I've got the right place to post. I've a test tomorrow and on the mock this question came up and I am confused about it.

The question is 'Make y the subject of: 6t2 = 5ay2 '

My working so far is divide both sides by 5a (6t2/5a=y2). Then do the square root of both sides (not sure if I do square root of just 6t2 or the whole side (6t2/5a)? Giving me 2.44t/5a = y.

My teacher has done y2=6t2/5a > y=square root of (6t2/5a) = t square root of (6/5a) I am most confused about this t square root. So his answer is y=t square root of 6/5a

Appreciate any help here, cheers

There is a general rule: \(\displaystyle \sqrt{a * b} \equiv \sqrt{a} * \sqrt{b}.\)

Are you familiar with it?

\(\displaystyle \therefore \sqrt{\dfrac{6t^2}{5a}} = \sqrt{t^2 * \dfrac{6}{5a}} = \sqrt{t^2} * \sqrt{\dfrac{6}{5a}} = t * \sqrt{\dfrac{6}{5a}}.\)

QED
 
Last edited:
Hi, I hope I've got the right place to post. I've a test tomorrow and on the mock this question came up and I am confused about it.

The question is 'Make y the subject of: 6t2 = 5ay2 '

My working so far is divide both sides by 5a (6t2/5a=y2). Then do the square root of both sides (not sure if I do square root of just 6t2 or the whole side (6t2/5a)? Giving me 2.44t/5a = y.

My teacher has done y2=6t2/5a > y=square root of (6t2/5a) = t square root of (6/5a) I am most confused about this t square root. So his answer is y=t square root of 6/5a

Appreciate any help here, cheers

If y2 = k (and k is positive), then y = +/- sqrt(k). So the correct answer to your problem is y = +/- tsqrt(6/(5a))
 
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