Core practice ← Back to lesson

Simplifying Radical Expressions: 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)

0 of 10 completed

Progress saved in this browser.

Problem 1 of 10

Problem 1 A squared input

Simplify $\sqrt{(x^2+1)^2}$ for real $x$ .
1. Hint 1
  The principal square root of a square is the absolute value of the squared expression.
2. Hint 2
  Determine whether $x^2+1$ can be negative.
Answer

$x^2+1$ .

Full solution

Taking the principal square root gives
$\sqrt{(x^2+1)^2}=|x^2+1|$

Since $x^2\ge0$ , the expression inside the bars is positive.
Therefore the simplified result is $x^2+1$ .

Answer

$x^2+1$ .

Key idea

An expression already known to be positive needs no absolute-value bars after an even root.
Problem 1: Mark complete
Problem 2 A missing coefficient

Find the positive number $c$ for which $\sqrt[3]{-189}=-c\sqrt[3]{7}$ .
1. Hint 1
  An odd root preserves the sign of its radicand.
2. Hint 2
  Write $189$ as a perfect cube times $7$ .
Answer

$c=3$ .

Full solution

The radicand is $(-27)\cdot7$ , so the odd-index product rule gives
$\sqrt[3]{-189}=-3\sqrt[3]{7}$

Thus $c=3$ .
Cubing the proposed right side gives
$-27\cdot7=-189.$

Answer

$c=3$ .

Key idea

An odd root allows a negative perfect-power factor to leave the radical with its sign.
Problem 2: Mark complete
Problem 3 An index to fill

For every positive $t$ , $\sqrt[14]{t^6}=\sqrt[k]{t^3}$ . Find the positive integer $k$ .
1. Hint 1
  Translate the two radicals into powers of $t$ .
2. Hint 2
  Reduce $6/14$ and match it with $3/k$ .
Answer

$k=7$ .

Full solution

The left side is $t^{6/14}$ , which is $t^{3/7}$ .
The right side is $t^{3/k}$ .
Matching these exponents for every positive input gives
$\frac3k=\frac37$

The required index is $7$ .
Both forms then give $t^{3/7}$ .

Answer

$k=7$ .

Key idea

For a positive base, a common factor of the radical index and the exponent reduces exactly as the fraction does.
Problem 3: Mark complete
Problem 4 A square panel

A square panel has area $175t^6$ square centimeters, where $t$ is a nonzero real calibration setting. Find its side length in simplest radical form. Explain how your expression handles a negative setting.
1. Hint 1
  The side length is the positive square root of the area.
2. Hint 2
  Look for perfect-square factors of $175$ and of the power of $t$ .
Answer

$5|t|^3\sqrt7$ cm; for $t<0$ this is $-5t^3\sqrt7$ cm, still a positive length.

Full solution

The area is positive because $t\ne0$ .
Take its principal square root and extract the perfect square factors.
$s=5|t|^3\sqrt7$

The bars preserve a positive length when $t$ is negative.
Squaring verifies the area.
$s^2=175t^6$

Answer

$5|t|^3\sqrt7$ cm; for $t<0$ this is $-5t^3\sqrt7$ cm, still a positive length.

Key idea

A geometric length from an even root must keep the positive sign when its parameter can be negative.
Problem 4: Mark complete
Problem 5 A divided input

For real $t\ne-1$ , simplify
$\sqrt[3]{\frac{(t-2)^3}{(t+1)^3}}$
Give the original restriction with your result.
1. Hint 1
  Odd roots preserve the signs of both numerator and denominator.
2. Hint 2
  The denominator is nonzero under the given restriction, so the quotient rule applies.
Answer

$\frac{t-2}{t+1}$ , with $t\ne-1$ .

Full solution

The radicand is the cube of $(t-2)/(t+1)$ , and this quotient exists because $t\ne-1$ .
The real cube root returns that quotient.
$\sqrt[3]{\frac{(t-2)^3}{(t+1)^3}}=\frac{t-2}{t+1}$

No bars are needed: negative quotients have negative cube roots.
Cubing the result recovers the radicand.

Answer

$\frac{t-2}{t+1}$ , with $t\ne-1$ .

Key idea

An odd root undoes the matching odd power of a defined quotient without changing its sign.
Problem 5: Mark complete
Problem 6 Two revisions of one expression

For $x>0$ , a student starts with $A=\sqrt[6]{64x^2}/\sqrt5$ , writes $B=2\sqrt[6]{x^2}/\sqrt5$ , and then writes $C=2\sqrt[3]{x}/\sqrt5$ . Verify both rewrites. Identify which condition for simplest radical form each rewrite addresses and which condition remains unmet in $C$ . Leave the denominator unchanged.
1. Hint 1
  Check the factor $64$ against the sixth-root index, then express the variable radical with a rational exponent.
2. Hint 2
  Use $64=2^6$ and reduce $2/6$ . Finally, inspect where the remaining radicals occur.
Answer

$A=B=C$ ; $A$ to $B$ : extracts a perfect sixth power; $B$ to $C$ : reduces the index; $C$ still has a radical denominator.

Full solution

Since $64=2^6$ and $x>0$ , the sixth-root product rule gives
$\sqrt[6]{64x^2}=2\sqrt[6]{x^2}$
Thus $A=B$ , removing a perfect sixth-power factor from the radicand.

For the positive variable,
$\sqrt[6]{x^2}=x^{2/6}$ $x^{2/6}=x^{1/3}$ $x^{1/3}=\sqrt[3]{x}$
This verifies $B=C$ and removes the common factor of the index and the exponent.

The denominator of $C$ is still $\sqrt5$ .
Thus the no-radical-denominator condition remains unmet, even though the two rewrites are valid.

Answer

$A=B=C$ ; $A$ to $B$ : extracts a perfect sixth power; $B$ to $C$ : reduces the index; $C$ still has a radical denominator.

Key idea

A valid sequence of radical rewrites may address some simplest-form conditions while leaving another unmet.
Problem 6: Mark complete
Problem 7 Two side lengths

Two square tiles have areas $207/64$ and $575/81$ square centimeters. Find both side lengths in simplest radical form, then determine which tile has the longer side.
1. Hint 1
  Take the positive square root of each area.
2. Hint 2
  After simplifying, compare the rational coefficients of the common radical.
Answer

$\frac{3\sqrt{23}}{8}$ cm and $\frac{5\sqrt{23}}{9}$ cm; the second tile has the longer side.

Full solution

The square roots of the positive denominators are $8$ and $9$ .
Extract square factors from the numerators: $207=9\cdot23$ , so
$s_1=\frac{3\sqrt{23}}{8}$

Since $575=25\cdot23$ , the other side is
$s_2=\frac{5\sqrt{23}}{9}$

The common factor $\sqrt{23}$ is positive.
Since $3/8=27/72$ and $5/9=40/72$ , the second coefficient and side are greater.
Squaring each length recovers its tile area.

Answer

$\frac{3\sqrt{23}}{8}$ cm and $\frac{5\sqrt{23}}{9}$ cm; the second tile has the longer side.

Key idea

Writing roots in simplest form can expose a common positive factor that makes comparison easy.
Problem 7: Mark complete
Problem 8 Different root signs

A student claims that two radical expressions with different indices must have different values. Give a counterexample using a square root and a fourth root, each with a positive integer radicand greater than $1$ .
1. Hint 1
  Two different radical signs can name the same principal value.
2. Hint 2
  Choose a positive integer, then use its square and fourth power as radicands.
Answer

For example, $\sqrt{16}=4$ and $\sqrt[4]{256}=4$ .

Full solution

Choose $4$ as the common positive value.
Its square and fourth power are the needed radicands.
$4^2=16$

The fourth power is
$4^4=256$

Thus the two different indices return the same number, disproving the claim.
Other examples satisfying the requested conditions also work.

Answer

For example, $\sqrt{16}=4$ and $\sqrt[4]{256}=4$ .

Key idea

The index alone does not determine a radical expression’s value.
Problem 8: Mark complete
Problem 9 A product claim

A student claims that $\sqrt{2t}\sqrt{8/t}=4$ is a valid identity over the real numbers exactly for $t>0$ . Is the claim correct? Justify both the identity and the stated domain.
1. Hint 1
  Check both radicands before combining them.
2. Hint 2
  Solve $2t\ge0$ and $8/t\ge0$ together, remembering the denominator restriction. Then apply the product rule on their common domain.
Answer

Yes; the identity holds exactly for $t>0$ .

Full solution

The expression $8/t$ requires $t\ne0$ .
Both square roots are real when $2t\ge0$ and $8/t\ge0$ , which together require $t>0$ .

On that domain the product rule applies.
$\sqrt{2t}\sqrt{8/t}=\sqrt{16}$ $\sqrt{16}=4$

At $t=0$ the quotient is undefined, and for negative $t$ the factors are not real.
Thus the stated domain is exact and the claim is correct.

Answer

Yes; the identity holds exactly for $t>0$ .

Key idea

A radical identity must preserve the domains of the original factors.
Problem 9: Mark complete
Problem 10 A sum under a root

Find two different positive integers $a$ and $b$ , neither a perfect square, with $a+b=196$ . Then use integer bounds, without decimals, to show that $\sqrt a+\sqrt b\ne\sqrt{a+b}$ for your pair.
1. Hint 1
  A sum can be a perfect square even when neither of its parts is.
2. Hint 2
  Bound each separate root between consecutive integers using nearby perfect squares.
Answer

One example: $a=83$ and $b=113$ . Then $\sqrt{a+b}=14$ , while $\sqrt{83}+\sqrt{113}>19$ . Other valid pairs are accepted.

Full solution

Take $a=83$ and $b=113$ .
Neither is a perfect square, since $81<83<100$ and $100<113<121$ , and
$83+113=196.$
So
$\sqrt{a+b}=14$

Since $83>81$ , $\sqrt{83}>9$ , and since $113>100$ ,
$\sqrt{113}>10.$
So
$\sqrt{83}+\sqrt{113}>19$

The sum of the separate roots is more than $19$ , and the root of the sum is $14$ , so they are not equal.
A square root does not split over addition.

Answer

One example: $a=83$ and $b=113$ . Then $\sqrt{a+b}=14$ , while $\sqrt{83}+\sqrt{113}>19$ . Other valid pairs are accepted.

Key idea

Square roots do not distribute over addition in general: for positive $a$ and $b$ , $\sqrt{a+b}<\sqrt a+\sqrt b$ .
Problem 10: Mark complete

← Back to lesson