Solve Equations with Nested Radicals
We've learnt a lot about solving equations including how to solve one step, two step and three step equations, as well as equations that include fractions. We've also looked at how to group pronumerals (ie. algebraic letters) when they are written on both sides of equation. And then, we've looked at various ways to solve quadratic equations.
We will now look at solving equations with nested radicals. Radicals are square roots, $\sqrt{x}$√x, cube roots, $\sqrt[3]{x}$3√x, etc. Nested radicals are when we have radicals nested under other radicals, giving expressions like $\sqrt{\sqrt[3]{4x+7}+6}$√3√4x+7+6.
Example
For example, we might want to solve an equation like this:
$\sqrt{\sqrt{x}+2}=3$√√x+2=3
The first thing you need to identify is what exactly is under the radical sign. Can you see here
- that only the $x$x is in the square root sign,
and that here
- that the expression $\sqrt{x}+2$√x+2 is under the square root sign.
As with most of our strategies to solve equations, the main technique we will use is to perform inverse operations. In this case we will reverse the radical sign by raising both sides of the equation to the relevant power. For example, if it is a square root, we square both sides, if it is a cube root, we cube both sides.
Applying this technique to our example, we get:
| $\left(\sqrt{\sqrt{x}+2}\right)^2$(√√x+2)2 | $=$= | $3^2$32 |
| $\sqrt{x}+2$√x+2 | $=$= | $9$9 |
Before we can apply the technique again, we need to isolate the $\sqrt{x}$√x by subtracting $2$2 from both sides of the equation:
$\sqrt{x}=7$√x=7
And now to remove the square root, we square both sides
| $\left(\sqrt{x}\right)^2$(√x)2 | $=$= | $7^2$72 |
| $x$x | $=$= | $49$49 |
As with all equations, it is a good habit to check that the solution works. This will be particularly important when we have equations with multiple solutions. Let's substitute $x=49$x=49 into the $LHS$LHS of the original equation, and check to see if it comes out with the value on the $RHS$RHS.
| $LHS$LHS | $=$= | $\sqrt{\sqrt{x}+2}$√√x+2 |
| $=$= | $\sqrt{\sqrt{49}+2}$√√49+2 | |
| $=$= | $\sqrt{7+2}$√7+2 | |
| $=$= | $\sqrt{9}$√9 | |
| $=$= | $3$3 | |
| $=$= | $RHS$RHS |
Another Example
When solving an equation of the form $3\sqrt{3\sqrt{3x}}=18$3√3√3x=18, be sure to isolate the $\sqrt{3\sqrt{3x}}$√3√3x part first, by dividing both sides by $3$3. We then alternate between applying the squaring technique, and isolating the relevant term by dividing by $3$3.
| $3\sqrt{3\sqrt{3x}}$3√3√3x | $=$= | $18$18 | remove the multiplication of $3$3 at the front |
| $\sqrt{3\sqrt{3x}}$√3√3x | $=$= | $6$6 | |
| $\left(\sqrt{3\sqrt{3x}}\right)^2$(√3√3x)2 | $=$= | $6^2$62 | square both sides |
| $3\sqrt{3x}$3√3x | $=$= | $36$36 | remove the multiplication of $3$3 at the front |
| $\sqrt{3x}$√3x | $=$= | $12$12 | |
| $\left(\sqrt{3x}\right)^2$(√3x)2 | $=$= | $12^2$122 | square both sides |
| $3x$3x | $=$= | $144$144 | |
| $x$x | $=$= | $48$48 | remove the multiplication of $3$3 at the front |
Before we finish, let's substitute $x=48$x=48 into the $LHS$LHS of the original equation, and check to see if it comes out with the value on the $RHS$RHS.
| $LHS$LHS | $=$= | $3\sqrt{3\sqrt{3x}}$3√3√3x |
| $=$= | $3\sqrt{3\sqrt{3\times48}}$3√3√3×48 | |
| $=$= | $3\sqrt{3\sqrt{144}}$3√3√144 | |
| $=$= | $3\sqrt{3\times12}$3√3×12 | |
| $=$= | $3\sqrt{36}$3√36 | |
| $=$= | $3\times6$3×6 | |
| $=$= | $18$18 | |
| $=$= | $RHS$RHS |
A Harder Example
Now we are going to look at a harder example, one which involves square roots and cube roots, and which has two possible solutions.
$\sqrt[3]{x+4+\sqrt{2x}}=2$3√x+4+√2x=2
We will use the same technique, of squaring, or cubing, both sides when the relevant expression is isolated. In the first step, we will cube both sides of the equation.
| $\sqrt[3]{x+4+\sqrt{2x}}$3√x+4+√2x | $=$= | $2$2 |
| $\left(\sqrt[3]{x+4+\sqrt{2x}}\right)^3$(3√x+4+√2x)3 | $=$= | $2^3$23 |
| $x+4+\sqrt{2x}$x+4+√2x | $=$= | $8$8 |
| $x+\sqrt{2x}$x+√2x | $=$= | $4$4 |
| $\sqrt{2x}$√2x | $=$= | $4-x$4−x |
| $\left(\sqrt{2x}\right)^2$(√2x)2 | $=$= | $(4-x)^2$(4−x)2 |
| $2x$2x | $=$= | $16-8x+x^2$16−8x+x2 |
| $16-8x+x^2$16−8x+x2 | $=$= | $2x$2x |
| $x^2-8x+16$x2−8x+16 | $=$= | $2x$2x |
| $x^2-10x+16$x2−10x+16 | $=$= | $0$0 |
We now have a quadratic equation, and we can use whichever technique we prefer. In this case, looking for two factors which multiply to make $16$16, and add to make $-10$−10, gives us $-2$−2 and $-8$−8. So I'm going to factorise the quadratic and then solve.
| $(x-2)(x-8)$(x−2)(x−8) | $=$= | $0$0 |
| $x-2=0$x−2=0 | or | $x-8=0$x−8=0 |
| $x=2$x=2 | or | $x=8$x=8 |
Just as we did with the previous examples, we want to check back with the original equation. In this case, we will need to check both results.
For $x=2$x=2
| $LHS$LHS | $=$= | $\sqrt[3]{x+4+\sqrt{2x}}$3√x+4+√2x |
| $=$= | $\sqrt[3]{2+4+\sqrt{2\times2}}$3√2+4+√2×2 | |
| $=$= | $\sqrt[3]{6+\sqrt{4}}$3√6+√4 | |
| $=$= | $\sqrt[3]{6+2}$3√6+2 | |
| $=$= | $\sqrt[3]{8}$3√8 | |
| $=$= | $2$2 | |
| $=$= | $RHS$RHS |
For $x=8$x=8
| $LHS$LHS | $=$= | $\sqrt[3]{x+4+\sqrt{2x}}$3√x+4+√2x |
| $=$= | $\sqrt[3]{8+4+\sqrt{2\times8}}$3√8+4+√2×8 | |
| $=$= | $\sqrt[3]{12+\sqrt{16}}$3√12+√16 | |
| $=$= | $\sqrt[3]{12+4}$3√12+4 | |
| $=$= | $\sqrt[3]{16}$3√16 | |
| $\ne$≠ | $2$2 | |
| $\ne$≠ | $RHS$RHS |
As we can see, $x=2$x=2 is a solution, and $x=8$x=8 is not a solution.
Always check all possible solutions, especially when there are $2$2 or more possible solutions.
Worked Examples
QUESTION 1
Solve $\sqrt{\sqrt{x-5}}=2$√√x−5=2.
Be sure to exclude any extraneous solutions.
QUESTION 2
Solve $\sqrt[3]{x+2+\sqrt{3x}}=2$3√x+2+√3x=2.
Be sure to exclude any extraneous solutions.
QUESTION 3
Solve $\sqrt{\sqrt{x^2+8x-1}+x-4}=3$√√x2+8x−1+x−4=3.
Be sure to exclude any extraneous solutions.