Radical terms are terms that have a root sign, such as a square root or cube root. Some equations contain radical terms and, in this lesson, we are going to look at how to solve these kinds of equations.
The process is basically the same as solving "regular" linear equations. We can still use inverse operations (that is, using the opposite operation) to solve equations. Remember, the inverse operation of a square root is squaring.
How does squaring "undo" a square root? Let's look at the general case of the $m$mth root.
The fractional exponent property states: $x^{\frac{1}{m}}=\sqrt[m]{x}$x1m=^{m}√x
The power of a power property states: $\left(x^m\right)^n=x^{mn}$(xm)n=xmn
So:
$\left(\sqrt[m]{x}\right)^m$(^{m}√x)m | $=$= | $\left(x^{\frac{1}{m}}\right)^m$(x1m)m |
$=$= | $x^{\frac{m}{m}}$xmm | |
$=$= | $x^1$x1 | |
$=$= | $x$x |
Solve the equation $3\sqrt{x}-5=-1$3√x−5=−1.
Think: We want to solve for $x$x, so we can use inverse operations to get it by itself.
Do:
$3\sqrt{x}-5$3√x−5 | $=$= | $-1$−1 | (State the original equation) |
$3\sqrt{x}$3√x | $=$= | $4$4 | (Add $5$5 to both sides) |
$\sqrt{x}$√x | $=$= | $\frac{4}{3}$43 | (Divide both sides by $3$3) |
$\left(\sqrt{x}\right)^2$(√x)2 | $=$= | $\left(\frac{4}{3}\right)^2$(43)2 | (Square both sides) |
$x$x | $=$= | $\frac{16}{9}$169 | (Simplify) |
Reflect: As with any equation, we can check our answer by substituting it back in.
$3\sqrt{x}-5$3√x−5 | $=$= | $-1$−1 | (State original equation) |
$3\sqrt{\frac{16}{9}}-5$3√169−5 | $=$= | $-1$−1 | (Substitute in $x=\frac{16}{9}$x=169) |
$3\times\frac{4}{3}-5$3×43−5 | $=$= | $-1$−1 | (Use order of operations, starting with the square root of $\frac{16}{9}$169) |
$4-5$4−5 | $=$= | $-1$−1 | (Next we multiply) |
$-1$−1 | $=$= | $-1$−1 | (Finally we subtract) |
We get a true statement, so $x=\frac{16}{9}$x=169 is a solution.
Remember our parent function $f\left(x\right)=\sqrt{x}$f(x)=√x. It had a domain of $x\ge0$x≥0 and a range of $y\ge0$y≥0. When we are solving equations involving square roots, we need to remember that on every step of our solution, we must have $x\ge0$x≥0 and $\sqrt{x}\ge0$√x≥0.
Let's look at some examples, where if we aren't careful we can end up with extraneous solutions. It is very important to check our solutions in the original equation.
Solve the equation $\sqrt{x+3}+10=4$√x+3+10=4.
Think: We want to solve for $x$x, so we can use inverse operations to get it by itself.
Do:
$\sqrt{x+3}+10$√x+3+10 | $=$= | $4$4 | (State the original equation) |
$\sqrt{x+3}$√x+3 | $=$= | $-6$−6 | (Subtract $10$10 from both sides) |
We should actually stop here and say no solution, as this would violate the range of $f\left(x\right)=\sqrt{x}$f(x)=√x, but let's see what happens if we keep going. | |||
$\left(\sqrt{x+3}\right)^2$(√x+3)2 | $=$= | $\left(-6\right)^2$(−6)2 | (Square both sides) |
$x+3$x+3 | $=$= | $36$36 | (Simplify) |
$x$x | $=$= | $33$33 | (Subtract $3$3 from both sides) |
Reflect: Since the second step was mathematically invalid, this is not an actual solution, but is extraneous. We can show this by substituting back in.
$\sqrt{x+3}+10$√x+3+10 | $=$= | $4$4 | (State original equation) |
$\sqrt{33+3}+10$√33+3+10 | $=$= | $4$4 | (Substitute in $x=33$x=33) |
$\sqrt{36}+10$√36+10 | $=$= | $4$4 | (Use order of operations, and simplify the expression in the grouping symbol) |
$6+10$6+10 | $=$= | $4$4 | (Next we perform the square root) |
$16$16 | $=$= | $4$4 | (Finally we add) |
We get a false statement, so there is no solution to this equation.
Solve the equation $\sqrt{2x}=x-4$√2x=x−4.
Think: We want to solve for $x$x, so we can use inverse operations to get it by itself.
Do:
$\sqrt{2x}$√2x | $=$= | $x-4$x−4 | (State the original equation) |
$\left(\sqrt{2x}\right)^2$(√2x)2 | $=$= | $\left(x-4\right)^2$(x−4)2 | (Square both sides) |
$2x$2x | $=$= | $x^2-8x+16$x2−8x+16 | (Distribute using a pattern) |
$0$0 | $=$= | $x^2-10x+16$x2−10x+16 | (Combine Like Terms) |
$0$0 | $=$= | $\left(x-8\right)\left(x-2\right)$(x−8)(x−2) | (Factor, if possible) |
$x=8$x=8 | and | $x=2$x=2 | (Use zero product property) |
Reflect: We need to check if these are valid or extraneous solutions. By substituting, we can see that only $x=8$x=8 is a valid solution. $x=2$x=2 is an extraneous solution. We can also confirm this by sketching graphs of the equations $y=\sqrt{2x}$y=√2x and $y=x-4$y=x−4.
Solve $\sqrt{x}=\sqrt{13}$√x=√13.
Solve the equation to find the value of $y$y.
$\sqrt{y}-6=1$√y−6=1
Solve $\sqrt{-5x+2}=\sqrt{-2x+8}$√−5x+2=√−2x+8.
Solve simple rational and radical equations in one variable, and give examples showing how extraneous solutions may arise.