# Proper method for solving quadratic equations with exponents

$(\sqrt {x^2-5x+6}+\sqrt{x^2-5x+4})^{x/2}$ + $(\sqrt {x^2-5x+6}-\sqrt{x^2-5x+4})^{x/2}$ = $2^{(x+4)/4}$

I have found out, by trial and error method, that $x=0$ and $x=4$ satisfy this equation. But is there a proper way to solve this equation and get the solutions?

• One way is to square both sides, noting that the LHS will have several terms become "nicer" upon doing so. Another is to multiply/divide by conjugates of one of the terms on the LHS and see if things simplify. – abiessu Oct 31 '13 at 14:17
• Note $2^{(x+4)/4} = 2\cdot2^{x/4} = 2\left(\sqrt{x^2-5x+6}-\sqrt{x^2-5x+4}\right)^{x/4}\left(\sqrt{x^2-5x+6}+\sqrt{x^2-5x+4}\right)^{x/4}$ – peterwhy Oct 31 '13 at 14:22

Let's start by putting $$y_1=\left(\sqrt {x^2-5x+6}+\sqrt{x^2-5x+4}\right)^{x/4}$$ and $$y_2=\left(\sqrt {x^2-5x+6}-\sqrt{x^2-5x+4}\right)^{x/4}.$$ Now, observe that \begin{align}2 &= (x^2-5x+6)-(x^2-5x+4)\\ &= \left(\sqrt {x^2-5x+6}+\sqrt{x^2-5x+4}\right)\left(\sqrt {x^2-5x+6}-\sqrt{x^2-5x+4}\right),\end{align} and so $$2^{(x+4)/4}=2\cdot 2^{x/4}=2y_1y_2.$$ Thus, the given equation becomes equivalent to solving $$y_1^2+y_2^2=2y_1y_2,$$ or $$(y_1-y_2)^2=0.$$ Hence, we need only solve the equation $$y_1=y_2,$$ so since both $y_1,y_2>0$ (why?), then we face the equivalent task of solving $$y_1^2=y_2^2.$$ As you've seen, $x=0$ readily does the trick, but if $x\ne 0,$ then the only way that we can have $a^x=b^x$ for $a,b>0$ is if $a=b.$ Hence, it remains only to solve $$\sqrt {x^2-5x+6}+\sqrt{x^2-5x+4}=\sqrt {x^2-5x+6}-\sqrt{x^2-5x+4},$$ which I leave to you.
Put $$a=(\sqrt {x^2-5x+6}+\sqrt{x^2-5x+4})^{x/2},\\ b=(\sqrt {x^2-5x+6}-\sqrt{x^2-5x+4})^{x/2}$$ Then $$a+b=2^{(x+4)/4},\\ a \cdot b=2^{x/2}$$ By solving it we get that $a=b,$ or $$(\sqrt {x^2-5x+6}+\sqrt{x^2-5x+4})^{x/2}=(\sqrt {x^2-5x+6}-\sqrt{x^2-5x+4})^{x/2}.$$ It is a simple equation.