# Make x the subject in each of the following...

• Dec 12th 2010, 06:18 PM
jgv115
Make x the subject in each of the following...
I got this question but I'm not really sure how one part of the solutions came up

$tx^2+4tx+10=t$

Make x the subject in each of the following and give the values of t for which real solution(s) to the equation can be found.

So I complete the square and get up to

$tx^2+4tx+10=t$

$\frac {tx^2+4tx+10-t}{t}=0$

$x^2+4x=\frac{-10}{t} +1$ I moved $\frac{10}{t}-1$ to the other side

$(x+2)^2 = \frac {-10}{t} +5$

$(x+2)^2 = \frac {5(-2+t)}{t}$

So I just take the square root of both sides and subtract -2.

But the solutions say $x=-2 \pm\frac {\sqrt{5t(-2+t)}}{t}$

How come my answer is different...
• Dec 12th 2010, 06:57 PM
Prove It
$\displaystyle t\,x^2 + 4t\,x + 10 = t$

$\displaystyle x^2 + 4x + \frac{10}{t} = 1$

$\displaystyle x^2 + 4x + 2^2 - 2^2 + \frac{10}{t} = 1$

$\displaystyle \left(x + 2\right)^2 + \frac{10 - 4t}{t} = \frac{t}{t}$

$\displaystyle \left(x + 2\right)^2 = \frac{5t - 10}{t}$

$\displaystyle x + 2 = \pm \sqrt{\frac{5t - 10}{t}}$

$\displaystyle x + 2 = \pm \sqrt{\frac{5t^2 - 10t}{t^2}}$

$\displaystyle x + 2 = \pm \frac{\sqrt{5t(t - 2)}}{t}$

$\displaystyle x = 2\pm \frac{\sqrt{5t(t-2)}}{t}$.

The solution the book gives (which I have posted) involves rationalising the denominator.
• Dec 12th 2010, 10:31 PM
jgv115
Ahh alright, if I get a question like this in an exam or something would I have to rationalise every time? Is it wrong if I don't?
• Dec 12th 2010, 10:40 PM
Prove It
Quote:

Originally Posted by jgv115
Ahh alright, if I get a question like this in an exam or something would I have to rationalise every time? Is it wrong if I don't?

Both answers are correct, but it's always best to have a rational denominator.