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Solving Radical Equations: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Two square roots

Solve $\sqrt{x^2-120}=\sqrt{7x}$ over the real numbers. Identify any extraneous candidate and explain its cause.
1. Hint 1
  Both square roots must exist and are then zero or positive.
2. Hint 2
  Squaring gives a quadratic with all terms on one side.
Answer

$x=15$ ; the candidate $x=-8$ is extraneous, because neither radical is real there.

Full solution

Squaring both sides gives
$x^2-120=7x$

Rearrange and factor.
$(x-15)(x+8)=0$

At $x=15$ , both original sides are $\sqrt{105}$ .
At $x=-8$ , both radicands are $-56$ , so neither side is a real number.
The candidate $-8$ solves the squared equation only because squaring widened the domain, and it is extraneous.

Answer

$x=15$ ; the candidate $x=-8$ is extraneous, because neither radical is real there.

Key idea

Squaring two radicals can admit a candidate at which neither original radical is real, so every candidate is checked in the original equation.
Problem 1: Mark complete
Problem 2 Undoing an odd power

Solve $\sqrt[3]{(2x-7)^3}=x+6$ over the real numbers.
1. Hint 1
  Recall how a real cube root acts on a perfect cube.
2. Hint 2
  Simplify the left side, then solve the resulting linear equation.
Answer

$x=13$ .

Full solution

A real cube root undoes a cube for every real input, so the left side is $2x-7$ :
$2x-7=x+6$

Thus $x=13$ .
Check: $2(13)-7=19$ and $13+6=19$ , and
$\sqrt[3]{19^3}=19.$

Answer

$x=13$ .

Key idea

A cube root undoes a cube for every real input.
Problem 2: Mark complete
Problem 3 A quotient equation

Solve $\dfrac{\sqrt{x+6}}{\sqrt{x+1}}=2$ over the real numbers.
1. Hint 1
  The denominator root must be real and nonzero.
2. Hint 2
  Multiply by the positive denominator, then square both sides.
Answer

$x=\frac23$ .

Full solution

The denominator requires $x>-1$ , which also keeps the numerator root real.
Multiplication gives
$\sqrt{x+6}=2\sqrt{x+1}$

Square the zero or positive sides.
$x+6=4x+4$

Solving gives $x=2/3$ .
At that input, the quotient squared is
$(20/3)/(5/3)=4.$
The quotient is positive, so it equals $2$ as required.

Answer

$x=\frac23$ .

Key idea

The denominator of a radical quotient adds a nonzero condition to the real-root conditions.
Problem 3: Mark complete
Problem 4 A meeting point

The graph shows $f(x)=\sqrt{(x-1)^2}$ and $g(x)=x+1$ . Find the coordinates of their intersection and verify them algebraically.

The graphs of f and g.

Text description of this figure

A grid with the x-axis from -4 to 5 and the y-axis from -3 to 7, equal unit scales, both ticked and labeled at every integer. The graph labeled f is a solid V: it comes down from the point (-4, 5) with slope -1 to its vertex on the x-axis at x equals 1, then rises with slope 1 to the point (5, 4). The graph labeled g is a dashed straight line of slope 1, running from the point (-4, -3) at the lower left to the point (5, 6) at the upper right. The two graphs cross once, on the left arm of the V; the crossing point is not marked or labeled.
1. Hint 1
  An intersection has equal outputs from the two functions.
2. Hint 2
  Require $x+1\ge0$ before squaring their equality.
Answer

$(0,1)$ .

Full solution

At an intersection, $x+1$ is zero or positive.
Square the equality to obtain
$(x-1)^2=(x+1)^2$

Expanding and canceling the square and constant terms gives
$-2x=2x$

Hence $x=0$ .
Both original outputs are $1$ , so the intersection is $(0,1)$ and the candidate satisfies the sign restriction.

Answer

$(0,1)$ .

Key idea

Algebraic checks confirm which graph intersections satisfy the original radical equation.
Problem 4: Mark complete
Problem 5 After two squarings

A student solves a radical equation by squaring both sides twice, rearranging between the two squarings. The final equation has exactly three real solutions: one satisfies the original equation and two do not. Could the original equation have any other real solutions? Explain.
1. Hint 1
  Follow a genuine solution of the original equation through each squaring step.
2. Hint 2
  Every solution of the final equation is on its list of three.
Answer

No; the original equation's only real solution is the one of the three that checks.

Full solution

Suppose $a$ solves the original equation.
At $a$ its two sides are equal real numbers, and equal real numbers have equal squares, so $a$ solves the equation after the first squaring.
Rearranging keeps it a solution.

The same argument carries $a$ through the second squaring, so $a$ solves the final equation and is one of its three solutions.

So every real solution of the original equation is among the three candidates.
Two of them fail the original, so the one that passes is its only real solution.

Answer

No; the original equation's only real solution is the one of the three that checks.

Key idea

Squaring can add solutions but never loses one, so checking the complete candidate list finds every solution.
Problem 5: Mark complete
Problem 6 A pair of distances

Solve $\sqrt{x+16}+\sqrt{9-x}=7$ over the real numbers.
1. Hint 1
  Both radicands must be zero or positive, which bounds the input.
2. Hint 2
  Squaring the sum leaves one product of square roots to isolate.
Answer

$x=-7$ or $x=0$ .

Full solution

The domain is
$-16\le x\le9.$
Squaring and adding the two radicands gives
$25+2\sqrt{(x+16)(9-x)}=49$

Isolate the radical and square again.
$(x+16)(9-x)=144$

Expand, cancel $144$ , and factor.
$x(x+7)=0$

The candidates are $0$ and $-7$ .
In the original equation, they give respectively $4+3=7$ and $3+4=7$ , so both are valid.

Answer

$x=-7$ or $x=0$ .

Key idea

Two squaring steps can produce several candidates, each of which must satisfy the original sum.
Problem 6: Mark complete
Problem 7 A calibrated reading

A device is used only at settings $t\ge6$ and reports $R(t)=\sqrt{t(t-6)}$ . Find every real setting that gives $R(t)=\sqrt7$ , applying the device restriction. Identify any discarded candidate and the reason it is discarded.
1. Hint 1
  The outputs are principal square roots, so compare their radicands.
2. Hint 2
  Solve the quadratic, then distinguish the equation’s domain from the allowed device settings.
Answer

$t=7$ ; discard $t=-1$ for violating the device restriction.

Full solution

Squaring the output condition gives
$t^2-6t=7$

The quadratic factors as
$(t-7)(t+1)=0$

Both $7$ and $-1$ make the original radicand $7$ , so both satisfy the radical equation itself.
However, the device accepts only $t\ge6$ .
Therefore the only permitted setting is $7$ , and $-1$ is rejected by the situation.

Answer

$t=7$ ; discard $t=-1$ for violating the device restriction.

Key idea

A model’s allowed settings may exclude a genuine solution of its algebraic equation.
Problem 7: Mark complete
Problem 8 Two original records

Two records contain the equations $\sqrt{(x+2)^2}=x+2$ and $\sqrt{(x+2)^2}=-(x+2)$ . Squaring either produces an identity. Give the actual solution set of each original equation and explain why they differ.
1. Hint 1
  The radical is the absolute value of $x+2$ .
2. Hint 2
  The two right sides impose opposite sign requirements.
Answer

First equation: $x\ge-2$ ; second equation: $x\le-2$ . The first requires $x+2\ge0$ ; the second requires $x+2\le0$ .

Full solution

The common left side is $|x+2|$ .
It equals $x+2$ exactly when $x+2\ge0$ , and equals $-(x+2)$ exactly when $x+2\le0$ .

At $x=-2$ , both equations hold because all sides are zero.
Squaring removes the sign information and gives the same identity in both cases.
$(x+2)^2=(x+2)^2$

That identity has more solutions than either original equation.

Answer

First equation: $x\ge-2$ ; second equation: $x\le-2$ . The first requires $x+2\ge0$ ; the second requires $x+2\le0$ .

Key idea

The same squared equation can come from originals with different sign requirements.
Problem 8: Mark complete
Problem 9 Cubing both sides

A student claims that cubing both sides of $\sqrt[3]{x-4}=(x-4)/4$ introduces no extra solutions. Decide whether that claim is correct, solve the equation, and check every resulting value.
1. Hint 1
  Ask whether two different real numbers can have the same cube.
2. Hint 2
  Let $u=x-4$ after cubing, then factor the resulting cubic.
Answer

The claim is correct; $x=-4,4,12$ .

Full solution

All expressions are defined for real $x$ , and cubing preserves equality in both directions.
With $u=x-4$ , the cubed equation is
$u=\frac{u^3}{64}$

Rearranging gives
$u(u-8)(u+8)=0$

Thus $u=-8,0,8$ , giving $x=-4,4,12$ .
The original equation has both sides equal to $-2$ , $0$ , and $2$ , respectively.
All candidates survive.

Answer

The claim is correct; $x=-4,4,12$ .

Key idea

Cubing real expressions preserves their equality in both directions.
Problem 9: Mark complete
Problem 10 An identity after squaring

A student squares $\sqrt{4-x}=-\sqrt{4-x}$ and concludes that every $x\le4$ solves the original equation. Find the actual solution set and explain what the squared identity lost.
1. Hint 1
  Both roots exist on the stated domain, but the two sides have opposite signs.
2. Hint 2
  A zero or positive number equals its negative only at zero.
Answer

$x=4$ ; the squared identity loses the requirement that $\sqrt{4-x}=0$ .

Full solution

The original equation requires $x\le4$ .
Its left side is zero or positive and its right side is zero or negative, so equality forces
$\sqrt{4-x}=0$

This gives $x=4$ , which makes both original sides zero.
For every smaller $x$ , the sides are nonzero opposites.
Squaring makes these opposites equal and therefore loses the condition that the root be zero.

Answer

$x=4$ ; the squared identity loses the requirement that $\sqrt{4-x}=0$ .

Key idea

A domain check does not recover the sign information lost by squaring.
Problem 10: Mark complete

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