# Thread: Assistance required with what seems a basic simplification.

1. ## Assistance required with what seems a basic simplification.

My expression is -3=[(-x*pi)/(sqrt(1-x^2))]

An online equation solver gave a solution of x=3/[sqrt(9+pi^2)] or x=.690621 but does not show the working.

I have spent hours on this but am positively stuck!

Can anyone help?

Regards.

2. Hello!
Use LaTeX in future for writing equations.

$-3=\frac{-\pi x}{\sqrt{1-x^2}}$

Hint:

$(-3)^2=(\frac{-\pi x}{\sqrt{1-x^2}})^2$

3. $\displaystyle -3 = \frac{-x\times \pi}{\sqrt{1-x^2}}$

$\displaystyle -3 \sqrt{1-x^2}= -x\times \pi$

$\displaystyle 3 \sqrt{1-x^2}= x\times \pi$

$\displaystyle 3^2 (1-x^2)= x^2\times \pi^2$

$\displaystyle 9-9x^2= x^2\times \pi^2$

$\displaystyle 9-9x^2- \pi^2x^2=0$

$\displaystyle 9-(9+ \pi^2)x^2 =0$

$\displaystyle (9+ \pi^2)x^2 =9$

$\displaystyle x^2 =\frac{9}{9+ \pi^2}$

$\displaystyle x =\sqrt{\frac{9}{9+ \pi^2}}$

4. Thank you very much for your replies - that one was seriously haunting me!

Regards.

5. I am curious to know - what rule tells me I cannot touch the terms under the square root?

To me sqrt(9+x^2) is 3+x but this is clearly not the case in this instance.

Some further clarification would be much appreciated.

Regards.

6. $\displaystyle \sqrt{9+x^2} \neq 3+x$

I'm not sure I follow what you are saying.

7. Obviously I am a bit rusty with my laws of powers etc. I thought sqrt(9+x^2) = sqrt(9) + sqrt(x^2) = 3 +x.

As I said - a bit rusty. Further reading necessary.

Regards.

8. You may be confused with

$\displaystyle \sqrt{a\times b } = \sqrt{a}\sqrt{b}$

$\displaystyle \sqrt{a+ b } \neq \sqrt{a}+\sqrt{b}$

9. In your second line did you mean sqrt{a+b }?

I apologise - not only am I rusty at powers but am new to Latex.

10. Generally speaking, you can only simplify a radical expression that is in factored form.

$\sqrt{9x^2}$ for example CAN be simplified to $3x$

The expression $\sqrt{x^2 + 9}$ on the other hand cannot be simplfied as-is.

This is especially apparent when you recall that $(a + b)^2 \neq a^2 + b^2$, because the square root would be the same as $(a + b)^{\frac{1}{2}}$, which by the same token, is not equal to $a^{\frac{1}{2}} + b^{\frac{1}{2}}$.

11. Great information. Thank you.

Makes me think of BOMDAS - brackets first, then powers.

Regards.