# Quadratic identities

## Homework Statement

By substituting y for 9x^2 solve 81x^4 - 63x^2 + 10 = 0

## The Attempt at a Solution

My attempt at a solution is:

y^2 - 7y + 10
(y-2)(y-5)

therefore y = 2 y = 5 Can someone double check this?

Cheers

## Answers and Replies

That's correct. Now solve for x by setting 9x2 equal to each solution of y.

Many thanks for the reply?

Is this to check the solution?

It just seemed to easy for some reason - maths is not my strong point, might I add.

Cheers

That's correct. Now solve for x by setting 9x2 equal to each solution of y.

Oh I see - are you then saying

9x^2 = 2
9x = +- Sqroot 2
x = + - Sqroot 2/9

9x^2 = 5
9x = +- Sqroot 5
x= + - Sqroot 5/9

Cheers

Mentallic
Homework Helper
Yes that is correct
You can even check for yourself by substituing those x values back into the original quadratic equation. You should find they work.

The only reason it asked you to substitute $$y=9x^2$$ into the equation $$81x^4 - 63x^2 + 10 = 0$$ is because it makes it more simple and easy to see how it should be solved.

Rather than substituing, you could've always factorized it as so:
$$81x^4-63x^2+10=(9x^2-2)(9x^2-5)=0$$

Oh I see - are you then saying

9x^2 = 2
9x = +- Sqroot 2
x = + - Sqroot 2/9

9x^2 = 5
9x = +- Sqroot 5
x= + - Sqroot 5/9

Cheers

√(ab) = √(a)√(b) ≠ a√(b) which is what you did between the first and second lines above.

You need to divide by 9 first, then take the square root of both sides.
$$9x^2 = 2 \rightarrow x^2 = \frac{2}{9} \rightarrow x = \pm\sqrt{\frac{2}{3}} = \pm\frac{\sqrt{2}}{3}$$

Or take the square root of both sides completely, then solve for x
$$9x^2 = 2 \rightarrow \sqrt{9x^2} = \sqrt{2} \rightarrow 3x = \pm\sqrt{2} \rightarrow x = \pm\frac{\sqrt{2}}{3}$$

Mentallic
Homework Helper
Oh when I skimmed through it I read sqroot 2/9 as $$\sqrt{\frac{2}{9}}$$. I didn't believe simplifying was top priority.
Thanks for spotting that Bohrok.

Hi there

Many thanks for both your assistance - it is greatly appreciated.

Cheers P