Lesson 5: Radical Equations

New Concept

A radical equation contains a variable under a radical, like $\sqrt{x + 3} = 4$. To solve, we'll isolate the radical and raise both sides to a power. We must also learn to spot and discard 'extraneous' solutions.

What’s next

Next, you’ll tackle interactive examples for solving these equations. Then, you'll use practice cards to master isolating radicals and checking for extraneous solutions.

Property

A radical equation is one in which the variable appears under a square root or other radical. We solve simple radical equations by raising both sides to the appropriate power. To do this, first isolate the radical expression on one side of the equation. Then, raise both sides to the power that matches the index of the radical.

Examples

• To solve $3\sqrt{x-4}=15$, first divide by 3 to get $\sqrt{x-4}=5$. Square both sides: $(\sqrt{x-4})^2 = 5^2$, which gives $x-4=25$, so $x=29$.
• To solve $\sqrt[3]{y+1}+6=9$, first subtract 6 to get $\sqrt[3]{y+1}=3$. Then cube both sides: $(\sqrt[3]{y+1})^3 = 3^3$, which gives $y+1=27$, so $y=26$.
• Solve $2\sqrt{3x-2}=10$. Isolate the radical: $\sqrt{3x-2}=5$. Square both sides: $3x-2=25$. Solve for x: $3x=27$, so $x=9$.

Explanation

Think of this as unwrapping a present; raising to a power is the inverse operation that undoes a root. Isolating the radical first ensures that this unwrapping process is clean and doesn't create a more complicated expression to solve.

Property

Whenever we raise both sides of an equation to an even power, it is possible to introduce false or extraneous solutions. For this reason, if we raise both sides of an equation to an even power, we should check each apparent solution in the original equation.

Caution: When squaring both sides, it is not correct to square each term separately, because $(a + b)^2 \neq a^2 + b^2$. Always isolate the radical first or square the entire side as a single binomial.

Examples

• The equation $\sqrt{x}=-7$ has no solution, as a principal root cannot be negative. Squaring gives $x=49$. Checking this, $\sqrt{49}=7 \neq -7$, so 49 is an extraneous solution.
• Solve $\sqrt{x+7}=x-5$. Squaring gives $x+7=(x-5)^2=x^2-10x+25$, which simplifies to $x^2-11x+18=0$, or $(x-2)(x-9)=0$. Possible solutions are $x=2$ and $x=9$. Checking $x=9$ works. Checking $x=2$ gives $\sqrt{9}=3$ and $2-5=-3$, which fails. Thus, $x=2$ is extraneous.
• Solve $x=\sqrt{2x+8}$. Squaring gives $x^2=2x+8$, or $x^2-2x-8=0$, which factors to $(x-4)(x+2)=0$. Possible solutions are $x=4$ and $x=-2$. Checking $x=4$ works. Checking $x=-2$ gives $-2=\sqrt{4}=2$, which fails. Thus, $x=-2$ is extraneous.

Property

We can also solve formulas involving radicals for one variable in terms of the others. The method is the same as for solving radical equations: isolate the radical that contains the desired variable, and then raise both sides of the formula to the appropriate power to eliminate the radical.

Examples

• Solve the formula for the radius of a cone, $r=\sqrt{\frac{3V}{\pi h}}$, for the height $h$. Square both sides: $r^2=\frac{3V}{\pi h}$. Multiply by $h$ and divide by $r^2$ to get $h=\frac{3V}{\pi r^2}$.
• Solve the formula for an object's velocity, $v = \sqrt{v_0^2 + 2ad}$, for the initial velocity $v_0$. Square both sides: $v^2 = v_0^2 + 2ad$. Subtract $2ad$: $v^2 - 2ad = v_0^2$. Take the square root: $v_0 = \pm\sqrt{v^2 - 2ad}$.
• Solve the formula $z = \sqrt[3]{x^2-y}$ for $y$. Cube both sides: $z^3=x^2-y$. Add $y$ to both sides: $z^3+y=x^2$. Subtract $z^3$ from both sides to get $y=x^2-z^3$.

Explanation

This is a key algebraic skill for rearranging scientific and geometric formulas. You are not finding a number, but rather expressing one variable in terms of others, using the same inverse operations you use to solve for a numerical answer.

Property

Sometimes we need to square both sides of an equation more than once in order to eliminate all the radicals. The general strategy is to first isolate the more complicated radical on one side of the equation. Square both sides, simplify, and then isolate the remaining radical. Finally, square both sides again to find the solution, and always check for extraneous roots.

Caution: We cannot solve a radical equation by squaring each term separately. An expression like $(\sqrt{x-7}+\sqrt{x})^2$ must be expanded as a binomial.

Examples

• Solve $\sqrt{x-5}+\sqrt{x}=5$. Isolate a radical: $\sqrt{x-5}=5-\sqrt{x}$. Square both sides: $x-5=(5-\sqrt{x})^2 = 25-10\sqrt{x}+x$. Isolate the remaining radical: $-30=-10\sqrt{x}$, so $3=\sqrt{x}$. Square again: $x=9$.
• Solve $\sqrt{y+8}-\sqrt{y}=2$. Isolate a radical: $\sqrt{y+8}=2+\sqrt{y}$. Square both sides: $y+8 = 4+4\sqrt{y}+y$. Simplify: $4=4\sqrt{y}$, so $1=\sqrt{y}$. Square again to get $y=1$.
• Solve $\sqrt{3x+4}=\sqrt{x-1}+3$. Square both sides: $3x+4=(\sqrt{x-1}+3)^2=x-1+6\sqrt{x-1}+9$. Simplify: $2x-4=6\sqrt{x-1}$, so $x-2=3\sqrt{x-1}$. Square again: $(x-2)^2=9(x-1)$, so $x^2-4x+4=9x-9$. This gives $x^2-13x+13=0$. This requires the quadratic formula for solutions.

Property

Roots of Powers.

If $n$ is odd, then $\sqrt[n]{a^n} = a$.

If $n$ is even, then $\sqrt[n]{a^n} = |a|$.