Finding the intersection of 2 functions

Question

I keep making mistakes with these kinds of questions when the algebra is a little harder.

$g(x)=x^2-4x+6$

$h(x)=3\sqrt x$

I see the answer is $1$ and $4$? How to reach this?

Thanks in advance.

Answer 1

With $x=t^2$, the equation becomes

$$t^4-4t^2-3t+6=0.$$

[Anyway, as $\sqrt x$ is a non-negative number, we must restrict to $t\ge0$.]

There are closed-form formulas to solve such quartic polynomials in the general cas, but they are tedious.

If you assume that the problem has been arranged so that the roots are integer, then it suffices to try the divisors of $6$, namely $\pm1,\pm2,\pm3$ and $\pm6$.

It turns out that $t=1$ and $t=2$ are solutions, so that the equation factors as

$$t^4-4t^2-3t+6=(t^2+at+b)(t-1)(t-2).$$

By long division, the quadratic factor is $t^2+3t+3$, which doesn't have real roots.