Find an impossibly found root algebraically in this equation

**SweatingBear** · April 29th 2012, 03:34 PM

I have this equation

x^{(x^2-3x)}=x^2

I've found all roots except x = -1.

Can somebody help me?

**Prove It** · April 29th 2012, 07:30 PM

Originally Posted by SweatingBear

I have this equation

x^{(x^2-3x)}=x^2

I've found all roots except x = -1.

Can somebody help me?

Since the bases are equal, the powers must be equal. So

$\displaystyle \begin{align*} x^{x^2 - 3x} &= x^2 \\ x^2 - 3x &= 2 \\ x^2 - 3x - 2 &= 0 \\ (x - 3)(x + 1) &= 0 \\ x - 3 = 0 \textrm{ or } x + 1 &= 0 \\ x = 3 \textrm{ or } x &= -1 \end{align*}$

**a tutor** · April 29th 2012, 11:12 PM

Originally Posted by Prove It

Since the bases are equal, the powers must be equal. So

$\displaystyle \begin{align*} x^{x^2 - 3x} &= x^2 \\ x^2 - 3x &= 2 \\ x^2 - 3x - 2 &= 0 \\ (x - 3)(x + 1) &= 0 \\ x - 3 = 0 \textrm{ or } x + 1 &= 0 \\ x = 3 \textrm{ or } x &= -1 \end{align*}$

erm, x=1 is an obvious solution. x=0 is worth considering but fails in this case. The other two roots are roots of x^2-3x-2=0. These are not -1 and 3.

**Prove It** · April 30th 2012, 03:16 AM

Originally Posted by a tutor

erm, x=1 is an obvious solution. x=0 is worth considering but fails in this case. The other two roots are roots of x^2-3x-2=0. These are not -1 and 3.

Oops, I don't know how I could have gotten my factorisation wrong, sorry...

$\displaystyle \begin{align*} x^2 - 3x - 2 &= x^2 - 3x + \left(-\frac{3}{2}\right)^2 - \left(-\frac{3}{2}\right)^2 - 2 \\ &= \left(x - \frac{3}{2}\right)^2 - \frac{9}{4} - \frac{8}{4} \\ &= \left(x - \frac{3}{2}\right)^2 - \frac{17}{4} \\ &= \left(x - \frac{3}{2}\right)^2 - \left(\frac{\sqrt{17}}{2}\right)^2 \\ &= \left(x - \frac{3}{2} - \frac{\sqrt{17}}{2}\right)\left(x - \frac{3}{2} + \frac{\sqrt{17}}{2}\right) \end{align*}$

**a tutor** · April 30th 2012, 04:24 AM

and I missed x=-1.

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