## Disguised Quadratics

Sometimes equations are constructed that reduce to a quadratic if a suitable substitution is made. We can then solve this reduced equation and through solving a second equation, determine the solutions of the original untransformed equation. Some care is needed to ensure we don't create addition (wrong) solutions.

## Disguised Quadratics 1

This is about the simplest version of a disguised quadratic. Let us make the substitution $x= t^2$.

In which case, our equation becomes $x^2 - 3x + 2 = 0$, this factorises to give $(x-1)(x+3) = 0$ and hence $x = 1$ or $x =-3$

Now recall that $x = t^2$ and so we have $t^2 = 1$ or $t^2 = -3$. These equations give $t = \pm 1$ or $t = \pm\sqrt{-3}$, given that we sought only real solutions, then our answers are $t= 1$ or $t= -1$

## Disguised Quadratics 2

Given that $x\in\mathbb{R}$, solve the following equation $$ 3(2^{2x}) - 3(2^x) - 6 = 0 $$

Let $t = 2^x$, then the equation above becomes $$ 3t^2 - 3t - 6 = 0 $$ This factorises to give $$ (3t-6)(t+1) = 0 $$ Hence, $t = 2$ or $t = -1$. Since t = 2^x, then we get $2^x = 2$, giving $x =1$ or $2^x = -1$, which gives no real solutions.