Math · College algebra · Worked example

Solve an equation with two square roots

Solve $\sqrt{2 x + 3} - \sqrt{x + 1} = 1$ .

\sqrt{2 x + 3} - \sqrt{x + 1} = 1

Isolate one root

Add $\sqrt{x + 1}$ to both sides. Isolating the more complicated root first keeps the next line simpler.

\sqrt{2 x + 3} = 1 + \sqrt{x + 1}

Square both sides

The left side becomes $2 x + 3$ . The right side is a binomial, ${(1 + \sqrt{x + 1})}^{2}$ $= 1 + 2 \sqrt{x + 1} + (x + 1)$ . A root is still there, in the middle term.

2 x + 3 = x + 2 + 2 \sqrt{x + 1}

Isolate the remaining root

Subtract $x + 2$ from both sides.

x + 1 = 2 \sqrt{x + 1}

Square again

The left side squares to $x^{2} + 2 x + 1$ . The right side squares to $4 (x + 1) = 4 x + 4$ .

x^{2} + 2 x + 1 = 4 x + 4

Solve the quadratic

Move every term to one side to get $x^{2} - 2 x - 3 = 0$ , which factors as $(x - 3) (x + 1) = 0$ . The candidates are $3$ and $- 1$ .

\begin{array}{l} x^{2} - 2 x - 3 = 0 \\ (x - 3) (x + 1) = 0 \end{array}

Check both candidates

Squaring twice could have added either one, so check both. At $x = 3$ , $\sqrt{9} - \sqrt{4} = 3 - 2 = 1$ . At $x = - 1$ , $\sqrt{1} - \sqrt{0} = 1 - 0 = 1$ . Both work, so neither is extraneous this time.

\begin{aligned} \sqrt{9} - \sqrt{4} & = 1 \\ \sqrt{1} - \sqrt{0} & = 1 \end{aligned}

Result

$x = 3$ or $x = - 1$ . Both satisfy the original equation.

x = 3 or x = - 1

Your turn

1. One like the example

Solve $\sqrt{x + 5} - \sqrt{x} = 1$ .

Show the answer and explanation

$x = 4$ .

Isolate $\sqrt{x + 5} = 1 + \sqrt{x}$ and square to get $x + 5 = 1 + 2 \sqrt{x} + x$ . The $x$ terms cancel, leaving $4 = 2 \sqrt{x}$ , so $\sqrt{x} = 2$ and $x = 4$ . Check: $\sqrt{9} - \sqrt{4} = 3 - 2 = 1$ . In a Math box every row reads ✓, the squaring row and the answer included.

\begin{array}{l} \sqrt{x + 5} - \sqrt{x} = 1 \\ \sqrt{x + 5} = 1 + \sqrt{x} \\ x + 5 = 1 + 2 \sqrt{x} + x \\ 4 = 2 \sqrt{x} \\ \sqrt{x} = 2 \\ x = 4 \end{array}

2. A variation

Solve $\sqrt{x + 3} + \sqrt{x - 2} = 5$ . This time the roots are added.

Show the answer and explanation

$x = 6$ .

Isolate $\sqrt{x + 3} = 5 - \sqrt{x - 2}$ and square. The right side is $25 - 10 \sqrt{x - 2} + (x - 2)$ , and the $x$ terms cancel to leave $10 \sqrt{x - 2} = 20$ . Then $\sqrt{x - 2} = 2$ and $x = 6$ . Check: $\sqrt{9} + \sqrt{4} = 3 + 2 = 5$ . As in the worked example, a Math box reads the first squaring row and the answer ✓, and the substitution is still a good check.

\begin{array}{l} \sqrt{x + 3} + \sqrt{x - 2} = 5 \\ \sqrt{x + 3} = 5 - \sqrt{x - 2} \\ x + 3 = 25 - 10 \sqrt{x - 2} + x - 2 \\ 10 \sqrt{x - 2} = 20 \\ \sqrt{x - 2} = 2 \\ x = 6 \end{array}

3. Somewhere new

Solve $\sqrt{x + 1} = \sqrt{2 x - 5}$ . Here each side is a root.

Show the answer and explanation

$x = 6$ .

With a root alone on each side, one squaring removes both, giving $x + 1 = 2 x - 5$ and $x = 6$ . Both radicands must be nonnegative, and at $x = 6$ they are $7$ and $7$ . Check: $\sqrt{7} = \sqrt{7}$ . Chalk reads this squaring row ✓, since it adds no solution here.

\begin{array}{l} \sqrt{x + 1} = \sqrt{2 x - 5} \\ x + 1 = 2 x - 5 \\ x = 6 \end{array}

Keep exploring

Open the steps in Math. The two squaring rows read ✓ because squaring added no solution here. Drop the cross term from the first one, $2 x + 3 = x + 2$ , and the box marks it ✗.

Return to the concept →

Sources and scope

Authored study material. Tool results depend on the stated inputs and model assumptions.

After reading

Study it, then try it

Practice the key terms as flashcards, then open the example in its tool, change a value and keep a useful result on your board.

Study the 5 key terms Check the steps in Math Open worked example on a board Checking after squaring in Math Reference

Your existing work stays on this device. Examples open as editable copies.

Isolate one root

Square both sides

Isolate the remaining root

Square again

Solve the quadratic

Check both candidates

Result

Your turn

1. One like the example

x=4x=4.

2. A variation

x=6x=6.

3. Somewhere new

x=6x=6.

Keep exploring

Study it, then try it

$x = 4$ .

$x = 6$ .

$x = 6$ .