Can anyone help with this as I am stumped:
1 + square root of x = square root of 3(x-1)
The answer is x=4 but I can't get it.
Also can you tell me how to do the square root symbol etc when posting a thread.
Thanks and Kind Regards
Larsson
Can anyone help with this as I am stumped:
1 + square root of x = square root of 3(x-1)
The answer is x=4 but I can't get it.
Also can you tell me how to do the square root symbol etc when posting a thread.
Thanks and Kind Regards
Larsson
Ok, i'll go through it
$\displaystyle 1+\sqrt{x}=\sqrt{3(x-1)}$
If we square both sides :
$\displaystyle \underbrace{(1+\sqrt{x})^2}_{1+x+2\sqrt{x}}=3(x-1)=3x-3$
$\displaystyle 1+x+2 \sqrt{x}=3x-3$
$\displaystyle 2 \sqrt{x}=2x-4$
$\displaystyle x-\sqrt{x}-2=0$
Substitute $\displaystyle u=\sqrt{x}$
$\displaystyle x=(\sqrt{x})^2=u^2$
So it's the same as solving for u (positive) in :
$\displaystyle u^2-u-2=0$
This is $\displaystyle (u-2)(u+1)=0$
So u is either equal to 2 or to -1.
But $\displaystyle u=\sqrt{x}>0$
So u=2.
Thus x=...