Equations reducible to quadratics

A ladder leans against a wall. How far from the wall must its foot be so that it reaches a given height? Pythagoras puts a square root in the equation. To get rid of the root we square both sides, but squaring can create fake solutions. This lesson shows how to square safely.

Why squaring needs a check

If $A=B$ then $A^2=B^2$. The reverse is false: $A^2=B^2$ only says $A=B$ or $A=-B$. For instance $x=-3$ is not a solution of $x=3$, yet squaring gives $x^2=9$, which accepts it. So after squaring we must either keep the right conditions or check every candidate in the original equation.

The equation $\sqrt{f(x)}=\sqrt{g(x)}$

Square both sides to get $f(x)=g(x)$. Then keep only the solutions for which $f(x)\ge 0$ (then $g(x)=f(x)\ge 0$ too).

The equation $\sqrt{f(x)}=g(x)$

A square root is never negative, so a solution needs $g(x)\ge 0$. Then squaring is safe: $$\sqrt{f(x)}=g(x)\iff\begin{cases}g(x)\ge 0\\ f(x)=g(x)^2.\end{cases}$$

A ladder problem

Key terms

Summary

• $\sqrt{f}=\sqrt{g}\iff f=g$ and $f\ge 0$.
• $\sqrt{f}=g\iff g\ge 0$ and $f=g^2$.
• Always check the candidates; squaring can add fake solutions.
• Real problems: translate to an equation, solve, then keep only meaningful answers.

Practice

Show solutions

1. $x+5=3x-1$ gives $x=3$. Check: $\sqrt8=\sqrt8$ ✓. $S=\{3\}$.
2. $x^2-4=3x$, so $x^2-3x-4=0$ and $x=4$ or $x=-1$. We need $3x\ge 0$: only $x=4$. Check: $\sqrt{12}=\sqrt{12}$ ✓. $S=\{4\}$.
3. $x-2=9$, so $x=11$. Check: $\sqrt9=3$ ✓.
4. Need $x\ge 0$. $x+2=x^2$, so $x^2-x-2=0$, $x=2$ or $x=-1$. Only $x=2$. Check: $\sqrt4=2$ ✓.
5. Need $x\ge 0$. $2x+3=x^2$, so $x^2-2x-3=0$, $x=3$ or $x=-1$. Only $x=3$. Check: $\sqrt9=3$ ✓.
6. Need $x\ge 3$. $x+3=x^2-6x+9$, so $x^2-7x+6=0$, $x=1$ or $x=6$. Only $x=6$. Check: $\sqrt9=3=6-3$ ✓.
7. Need $x\ge -4$. $2-x=x^2+8x+16$, so $x^2+9x+14=0$, $x=-2$ or $x=-7$. Only $x=-2$. Check: $\sqrt4=2=-2+4$ ✓.
8. $x^2-2x=x+4$, so $x^2-3x-4=0$ and $x=4$ or $x=-1$. Check: for $x=4$ both sides are $\sqrt8$ ✓; for $x=-1$ both sides are $\sqrt3$ ✓. $S=\{-1, 4\}$.
9. Need $x\ge 1$. $2x^2-3x-5=x^2-2x+1$, so $x^2-x-6=0$, $x=3$ or $x=-2$. Only $x=3$. Check: $\sqrt{18-9-5}=\sqrt4=2=3-1$ ✓.
10. $\sqrt{169-d^2}=d+7$. Squaring: $169-d^2=d^2+14d+49$, so $d^2+7d-60=0$ and $d=5$ or $d=-12$. So $d=5$ m and the height is $12$ m. Check: $\sqrt{169-25}=12=5+7$ ✓.