Solve Equations with Square Roots

By the end of this section, you will be able to: solve radical equations, and use square roots in applications.

In this section we solve equations that have the variable in the radicand of a square root. Equations of this type are called radical equations.

Radical equation. An equation in which the variable is in the radicand of a square root is called a radical equation.

Solve Radical Equations

As usual, whatever we do to one side of an equation we must do to the other. Since squaring a quantity and taking a square root are “opposite” operations, we square both sides to remove the radical sign and free the variable inside.

But remember that $\sqrt{a}$ means the principal (nonnegative) square root, so $\sqrt{a} \ge 0$ always. When we square both sides we may get an algebraic solution that would make $\sqrt{a}$ negative. That value is not a solution of the original radical equation — it is an extraneous solution, exactly as we saw with rational equations. Because squaring can introduce these, we must check every algebraic solution in the original equation.

Example. For the equation $\sqrt{x+2} = x$ , is $x = 2$ a solution? Is $x = -1$ ?

Substitute each value into the original equation.

\begin{array}{rcl} \sqrt{2+2} &\overset{?}{=}& 2 \\[4pt] \sqrt{4} &\overset{?}{=}& 2 \\[4pt] 2 &=& 2 \; \checkmark \end{array}

So $x = 2$ is a solution. Now test $x = -1$ .

\begin{array}{rcl} \sqrt{-1+2} &\overset{?}{=}& -1 \\[4pt] \sqrt{1} &\overset{?}{=}& -1 \\[4pt] 1 &\ne& -1 \end{array}

So $x = -1$ is not a solution — it is an extraneous solution to the equation.

Now we see how to solve a radical equation. Our strategy uses the relation between taking a square root and squaring: for $a \ge 0$ , $\left(\sqrt{a}\right)^2 = a$ .

Example. Solve: $\sqrt{2x-1} = 7$ .

\begin{array}{lrcl} & \sqrt{2x-1} &=& 7 \\[4pt] \text{The radical is already isolated. Square both sides.} & \left(\sqrt{2x-1}\right)^2 &=& (7)^2 \\[4pt] \text{Solve the new equation.} & 2x-1 &=& 49 \\[4pt] & 2x &=& 50 \\[4pt] & x &=& 25 \end{array}

Check $x = 25$ in the original equation.

\begin{array}{rcl} \sqrt{2(25)-1} &\overset{?}{=}& 7 \\[4pt] \sqrt{49} &\overset{?}{=}& 7 \\[4pt] 7 &=& 7 \; \checkmark \end{array}

The solution is $x = 25$ .

Solve a radical equation.

Isolate the radical on one side of the equation.
Square both sides of the equation.
Solve the new equation.
Check the answer.

Solve: $\sqrt{3x-5} = 5$ . (Enter the value of $x$ .)

When the radical is not alone, isolate it first, then square.

Example. Solve: $\sqrt{5n-4} - 9 = 0$ .

\begin{array}{lrcl} & \sqrt{5n-4} - 9 &=& 0 \\[4pt] \text{Add } 9 \text{ to both sides to isolate the radical.} & \sqrt{5n-4} &=& 9 \\[4pt] \text{Square both sides.} & \left(\sqrt{5n-4}\right)^2 &=& (9)^2 \\[4pt] \text{Solve the new equation.} & 5n-4 &=& 81 \\[4pt] & 5n &=& 85 \\[4pt] & n &=& 17 \end{array}

Checking $n = 17$ gives $\sqrt{81} - 9 = 9 - 9 = 0$ , so $0 = 0\ \checkmark$ . The solution is $n = 17$ .

Solve: $\sqrt{3m+2} - 5 = 0$ . (Enter the value of $m$ .)

When there is no solution

When we use a radical sign, we mean the principal (nonnegative) root. If an isolated radical equals a negative number, the equation has no solution.

Example. Solve: $\sqrt{9k-2} + 1 = 0$ .

\begin{array}{lrcl} & \sqrt{9k-2} + 1 &=& 0 \\[4pt] \text{Subtract } 1 \text{ from both sides to isolate the radical.} & \sqrt{9k-2} &=& -1 \end{array}

Since a principal square root can never equal a negative number, the equation has no solution.

Solve: $\sqrt{2r-3} + 5 = 0$ . After isolating the radical, you get $\sqrt{2r-3} = -5$ . Which is correct?

A binomial on one side

If one side of the equation is a binomial after squaring, use the binomial squares formulas — and don’t forget the middle term:

(a+b)^2 = a^2 + 2ab + b^2 \qquad (a-b)^2 = a^2 - 2ab + b^2

Example. Solve: $\sqrt{p-1} + 1 = p$ .

Isolate the radical, then square. The right side is a binomial, so squaring it gives three terms.

\begin{array}{lrcl} & \sqrt{p-1} + 1 &=& p \\[4pt] \text{Subtract } 1 \text{ to isolate the radical.} & \sqrt{p-1} &=& p - 1 \\[4pt] \text{Square both sides.} & \left(\sqrt{p-1}\right)^2 &=& (p-1)^2 \\[4pt] \text{Simplify — a quadratic. Get zero on one side.} & 0 &=& p^2 - 3p + 2 \\[4pt] \text{Factor.} & 0 &=& (p-1)(p-2) \\[4pt] \text{Zero product property.} & p = 1 &\text{or}& p = 2 \end{array}

Both check: for $p = 1$ , $\sqrt{0} + 1 = 1\ \checkmark$ ; for $p = 2$ , $\sqrt{1} + 1 = 2\ \checkmark$ . The solutions are $p = 1$ and $p = 2$ .

Squaring a binomial can also introduce an extraneous solution, so the check step is essential.

Example (extraneous root discarded). Solve: $\sqrt{r+4} - r + 2 = 0$ .

\begin{array}{lrcl} & \sqrt{r+4} - r + 2 &=& 0 \\[4pt] \text{Isolate the radical.} & \sqrt{r+4} &=& r - 2 \\[4pt] \text{Square both sides.} & \left(\sqrt{r+4}\right)^2 &=& (r-2)^2 \\[4pt] \text{Simplify — a quadratic. Get zero on one side.} & 0 &=& r^2 - 5r \\[4pt] \text{Factor.} & 0 &=& r(r-5) \\[4pt] \text{Zero product property.} & r = 0 &\text{or}& r = 5 \end{array}

Checking $r = 5$ gives $\sqrt{9} - 5 + 2 = 3 - 3 = 0\ \checkmark$ . But checking $r = 0$ gives $\sqrt{4} + 2 = 4 \ne 0$ , so $r = 0$ is an extraneous solution. The only solution is $r = 5$ .

Solve: $\sqrt{m+9} - m + 3 = 0$ . It has two algebraic solutions but only one checks. (Enter the valid value of $m$ .)

A coefficient in front of the radical

When a coefficient multiplies the radical, isolate the whole radical term first; squaring then squares the coefficient too.

Example. Solve: $3\sqrt{3x-5} - 8 = 4$ .

\begin{array}{lrcl} & 3\sqrt{3x-5} - 8 &=& 4 \\[4pt] \text{Isolate the radical term.} & 3\sqrt{3x-5} &=& 12 \\[4pt] \text{Square both sides — the coefficient squares too.} & \left(3\sqrt{3x-5}\right)^2 &=& (12)^2 \\[4pt] \text{Simplify.} & 9(3x-5) &=& 144 \\[4pt] \text{Distribute, then solve.} & 27x - 45 &=& 144 \\[4pt] & 27x &=& 189 \\[4pt] & x &=& 7 \end{array}

Checking $x = 7$ gives $3\sqrt{16} - 8 = 3(4) - 8 = 4\ \checkmark$ . The solution is $x = 7$ .

Solve: $2\sqrt{4a+2} - 16 = 16$ . (Enter the value of $a$ .)

A radical on each side

When there is a radical on each side and both are isolated, square both sides directly.

Example. Solve: $\sqrt{4z-3} = \sqrt{3z+2}$ .

Both radicals are isolated, so square both sides. The squares undo the radicals, leaving a linear equation.

\begin{array}{lrcl} & \sqrt{4z-3} &=& \sqrt{3z+2} \\[4pt] \text{Square both sides.} & \left(\sqrt{4z-3}\right)^2 &=& \left(\sqrt{3z+2}\right)^2 \\[4pt] \text{Solve the new equation.} & 4z-3 &=& 3z+2 \\[4pt] & z-3 &=& 2 \\[4pt] & z &=& 5 \end{array}

Checking $z = 5$ gives $\sqrt{17} = \sqrt{17}\ \checkmark$ . The solution is $z = 5$ .

Squaring twice

Sometimes, after squaring both sides, a radical still remains. When that happens, repeat the procedure: isolate the remaining radical and square again.

Example. Solve: $\sqrt{m} + 1 = \sqrt{m+9}$ .

\begin{array}{lrcl} & \sqrt{m} + 1 &=& \sqrt{m+9} \\[4pt] \text{Square both sides — be careful with the binomial.} & \left(\sqrt{m}+1\right)^2 &=& \left(\sqrt{m+9}\right)^2 \\[4pt] \text{Simplify. A radical still remains.} & m + 2\sqrt{m} + 1 &=& m + 9 \\[4pt] \text{Isolate the remaining radical.} & 2\sqrt{m} &=& 8 \\[4pt] \text{Square both sides again.} & \left(2\sqrt{m}\right)^2 &=& (8)^2 \\[4pt] \text{Solve.} & 4m &=& 64 \\[4pt] & m &=& 16 \end{array}

Checking $m = 16$ gives $\sqrt{16} + 1 = 5 = \sqrt{25}\ \checkmark$ . The solution is $m = 16$ .

Solve: $\sqrt{x} + 2 = \sqrt{x+16}$ . (Enter the value of $x$ .)

Use Square Roots in Applications

Formulas that include square roots appear across many disciplines. We solve these applications with the same plan we used for geometry problems: read, identify, name, translate to a formula, solve, check, and answer with a sentence.

Area of a square

A square is a rectangle whose length and width are equal, so a square with side $s$ has area $A = s^2$ . Solving that formula for the side gives a square root:

s = \sqrt{A}

Example. Mike and Lychelle want to make a square patio with an area of $200$ square feet. Use $s = \sqrt{A}$ to find the length of each side, rounded to the nearest tenth of a foot.

Substitute $A = 200$ :

s = \sqrt{200} = 14.14213\ldots \approx 14.1

Check: $14.1^2 = 198.81 \approx 200\ \checkmark$ (close, since we rounded). Each side of the patio should be about $14.1$ feet.

Katie wants a square lawn with an area of $370$ square feet. Use $s = \sqrt{A}$ to find the side length, rounded to the nearest tenth of a foot.

Falling objects

Another application involves gravity.

Falling objects. On Earth, if an object is dropped from a height of $h$ feet, the time in seconds it takes to reach the ground is

t = \frac{\sqrt{h}}{4}.

Example. Christy dropped her sunglasses from a bridge $400$ feet above a river. Use $t = \tfrac{\sqrt{h}}{4}$ to find how many seconds it took them to reach the river.

Substitute $h = 400$ :

t = \frac{\sqrt{400}}{4} = \frac{20}{4} = 5

Check: $\tfrac{\sqrt{400}}{4} = \tfrac{20}{4} = 5\ \checkmark$ . It took $5$ seconds for the sunglasses to reach the water.

A helicopter dropped a rescue package from a height of $1{,}296$ feet. Use $t = \tfrac{\sqrt{h}}{4}$ to find the number of seconds it took the package to reach the ground.

Skid marks and speed

Police officers measure the length of skid marks to estimate how fast a car was going before braking.

Skid marks and speed of a car. If the length of the skid marks is $d$ feet, then the speed $s$ of the car in miles per hour before braking is

s = \sqrt{24d}.

Example. After a car accident, the skid marks for one car measured $190$ feet. Use $s = \sqrt{24d}$ to find the speed of the car before braking, rounded to the nearest tenth.

Substitute $d = 190$ :

s = \sqrt{24(190)} = \sqrt{4560} = 67.52777\ldots \approx 67.5

Is $67.5$ mph a reasonable speed? Yes. The speed of the car was approximately $67.5$ miles per hour.

The skid marks of a car measured $76$ feet. Use $s = \sqrt{24d}$ to find the speed of the car before braking, rounded to the nearest tenth of a mile per hour.

Key terms

radical equation — an equation in which the variable is in the radicand of a square root. principal square root — the nonnegative square root of a number; $\sqrt{a} \ge 0$ always. extraneous solution — an algebraic solution obtained by squaring that does not satisfy the original radical equation (it would make a principal square root negative); it must be discarded, which is why every solution of a radical equation must be checked.

This section is adapted from Elementary Algebra 2e, 9.6 Solve Equations with Square Roots by Lynn Marecek, MaryAnne Anthony-Smith, and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: condensed the worked examples into aligned step tables and prose, recast the “How To” procedures as callouts, folded the no-solution and discarded-root cases into subsections, and summarized the seven-step application solutions; omitted the Be Prepared quiz, Self Check checklist, media links, and end-of-section exercises; and converted the practice problems (“Try Its”) into interactive exercises with instant feedback.

Divide Square Roots Higher Roots