- #1

- 107

- 0

## Homework Statement

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

## Homework Equations

## 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

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter zebra1707
- Start date

- #1

- 107

- 0

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

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

- #2

- 867

- 0

That's correct. Now solve for x by setting 9x^{2} equal to each solution of y.

- #3

- 107

- 0

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 9x^{2}equal to each solution of y.

- #4

- 107

- 0

9x^2 = 2

9x = +- Sqroot 2

x = + - Sqroot 2/9

9x^2 = 5

9x = +- Sqroot 5

x= + - Sqroot 5/9

Cheers

- #5

Mentallic

Homework Helper

- 3,798

- 94

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 [tex]y=9x^2[/tex] into the equation [tex]81x^4 - 63x^2 + 10 = 0[/tex] 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:

[tex]81x^4-63x^2+10=(9x^2-2)(9x^2-5)=0[/tex]

- #6

- 867

- 0

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.

[tex]9x^2 = 2 \rightarrow x^2 = \frac{2}{9} \rightarrow x = \pm\sqrt{\frac{2}{3}} = \pm\frac{\sqrt{2}}{3}[/tex]

Or take the square root of both sides completely, then solve for x

[tex]9x^2 = 2 \rightarrow \sqrt{9x^2} = \sqrt{2} \rightarrow 3x = \pm\sqrt{2} \rightarrow x = \pm\frac{\sqrt{2}}{3}[/tex]

- #7

Mentallic

Homework Helper

- 3,798

- 94

Thanks for spotting that Bohrok.

- #8

- 107

- 0

Hi there

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

Cheers P

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

Cheers P

Share: