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Operations with Radicals: 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 A coefficient slot

Find the real number $c$ for which $c\sqrt{13}+2\sqrt{13}=9\sqrt{13}$ .
1. Hint 1
  All the terms share the same nonzero radical factor.
2. Hint 2
  Divide by $\sqrt{13}$ and solve the resulting equation.
Answer

$c=7$ .

Full solution

Since $\sqrt{13}>0$ , divide through by it.
$c+2=9$

Subtract $2$ to obtain $c=7$ .
The coefficients $7$ and $2$ then add to $9$ .

Answer

$c=7$ .

Key idea

Like radical terms combine through their coefficients.
Problem 1: Mark complete
Problem 2 A product over a radical

Simplify $\dfrac{(6\sqrt{38})(5\sqrt{57})}{15\sqrt6}$ exactly.
1. Hint 1
  Handle the coefficients and the radicands separately, for the multiplication and for the division.
2. Hint 2
  Factor the radicands before multiplying them: $38=2\cdot19$ and $57=3\cdot19$ .
Answer

$38$ .

Full solution

The coefficients give
$\frac{6\cdot5}{15}=2.$
The radicands give $\frac{38\cdot57}{6}$ , and since $38\cdot57=2\cdot3\cdot19^2$ ,
$\frac{38\cdot57}6=19^2$

All radicands are positive, so the product and quotient rules for square roots apply, and the value is
$2\sqrt{19^2}=38$

Answer

$38$ .

Key idea

For real radicals of one index, with nonnegative radicands when the index is even and a nonzero divisor, multiplying and dividing handle the coefficients and the radicands separately.
Problem 2: Mark complete
Problem 3 A product entry

Expand $\sqrt2(1+\sqrt2+\sqrt3)$ into a sum of terms in simplest radical form.
1. Hint 1
  Distribute the factor to every term inside the parentheses.
2. Hint 2
  One product squares $\sqrt2$ ; another combines the positive radicands $2$ and $3$ .
Answer

$2+\sqrt2+\sqrt6$ .

Full solution

The products are $\sqrt2$ , $2$ , and $\sqrt6$ .
Thus
$\begin{aligned}&\sqrt2(1+\sqrt2+\sqrt3)\\&=2+\sqrt2+\sqrt6\end{aligned}$

The remaining radical terms have different radicands and are already in simplest form.

Answer

$2+\sqrt2+\sqrt6$ .

Key idea

Distribution handles each term before like radicals are collected.
Problem 3: Mark complete
Problem 4 The remaining rod

A rod is $5\sqrt{17}$ cm long. Two pieces, each $\sqrt{68}$ cm long, are cut from it. Ignoring the cutting width, find the remaining length exactly.
1. Hint 1
  Simplify the piece length before subtracting it twice.
2. Hint 2
  $68=4\cdot17$ , so each piece is a multiple of $\sqrt{17}$ .
Answer

$\sqrt{17}$ cm.

Full solution

Each piece has length $2\sqrt{17}$ cm.
Together the pieces use $4\sqrt{17}$ cm.
The remaining length is
$5\sqrt{17}-4\sqrt{17}=\sqrt{17}$

The removed and remaining lengths add back to the original rod length.

Answer

$\sqrt{17}$ cm.

Key idea

Simplifying radical lengths exposes the common unit needed for addition and subtraction.
Problem 4: Mark complete
Problem 5 A rectangular card

A rectangular card has side lengths $3+\sqrt2$ cm and $3-\sqrt2$ cm. Find its exact area and perimeter.
1. Hint 1
  Area multiplies the side lengths, while perimeter adds two copies of each.
2. Hint 2
  In the area product, the two mixed terms cancel.
Answer

Area: $7$ square centimeters; perimeter: $12$ cm.

Full solution

Both sides are positive since $\sqrt2<3$ .
The area is a difference of squares.
$A=9-2$

This gives $7$ square centimeters.
The sum of the side lengths is $6$ cm, so
$P=12$

The radical terms cancel in the sum as well as in the area product.

Answer

Area: $7$ square centimeters; perimeter: $12$ cm.

Key idea

A matching radical sum and difference can produce rational geometric measurements.
Problem 5: Mark complete
Problem 6 A scale formula

A scale factor is $L=\sqrt[4]{x^3}/\sqrt[6]{x}$ for $x>0$ . Express $L$ as one power of $x$ , then find its value at $x=4096$ .
1. Hint 1
  Translate the two indices into exponents before dividing.
2. Hint 2
  Subtract $1/6$ from $3/4$ using denominator $12$ .
Answer

$L=x^{7/12}$ ; at $x=4096$ , $L=128$ .

Full solution

With positive $x$ , exponent laws apply and the denominator is nonzero.
$\frac34-\frac16=\frac7{12}$

Thus
$L=x^{7/12}.$
Since $4096=2^{12}$ , its twelfth root is $2$ .
$L=128$

The original numerator is $512$ and denominator is $4$ , confirming $128$ .

Answer

$L=x^{7/12}$ ; at $x=4096$ , $L=128$ .

Key idea

When radicals are powers of the same positive base, dividing them subtracts their rational exponents, over a common denominator if the indices differ.
Problem 6: Mark complete
Problem 7 A missing integer

An integer $k$ is chosen so that $(2+\sqrt{59})(k-3\sqrt{59})$ is rational. Find $k$ and the resulting product.
1. Hint 1
  Expand and identify the coefficient of the irrational term.
2. Hint 2
  Because $k$ is an integer, the product is rational exactly when that radical coefficient is zero.
Answer

$k=6$ ; the product is $-165$ .

Full solution

Expansion gives
$2k-177+(k-6)\sqrt{59}$

The expression $2k-177$ is rational.
Since $\sqrt{59}$ is irrational, the other term is rational only when $k-6=0$ .
Therefore $k=6$ , and the product is
$2k-177=-165$

The second factor is then three times the conjugate of the first, so the radical terms cancel.

Answer

$k=6$ ; the product is $-165$ .

Key idea

Choosing coefficients to cancel the irrational term can make a radical product rational.
Problem 7: Mark complete
Problem 8 Two equal halves

Split the four numbers $\sqrt{116}$ , $\sqrt{261}$ , $-\sqrt{29}$ and $-2\sqrt{29}$ into two pairs with equal sums, using each number once. Give the pairs and their common sum.
1. Hint 1
  Simplify every radical first, to see whether the four numbers are like radicals.
2. Hint 2
  The four numbers have a total, and each pair must carry exactly half of it.
Answer

$(\sqrt{116},-\sqrt{29})$ and $(\sqrt{261},-2\sqrt{29})$ , each with sum $\sqrt{29}$ .

Full solution

Since $116=4\cdot29$ and $261=9\cdot29$ , the four numbers are $2\sqrt{29}$ , $3\sqrt{29}$ , $-\sqrt{29}$ and $-2\sqrt{29}$ : like radicals with coefficients $2$ , $3$ , $-1$ and $-2$ .

Their total is $2\sqrt{29}$ , so each pair must sum to $\sqrt{29}$ , which means coefficients summing to $1$ .
The only such split is $2+(-1)$ and $3+(-2)$ .

So the pairs are $\sqrt{116}$ with $-\sqrt{29}$ and $\sqrt{261}$ with $-2\sqrt{29}$ , and the other two ways to pair the numbers give sums $0$ and $2\sqrt{29}$ , or $5\sqrt{29}$ and $-3\sqrt{29}$ .

Answer

$(\sqrt{116},-\sqrt{29})$ and $(\sqrt{261},-2\sqrt{29})$ , each with sum $\sqrt{29}$ .

Key idea

Simplifying first shows which radicals are alike, and like radicals add by their coefficients.
Problem 8: Mark complete
Problem 9 The sign of a product

For $a,b\ge0$ , a student claims $(\sqrt a+\sqrt b)(\sqrt a-\sqrt b)$ is zero or positive exactly when $a\ge b$ . Is the claim true? Justify it.
1. Hint 1
  The matching sum and difference eliminate the mixed radical terms.
2. Hint 2
  The remaining expression compares the two radicands directly.
Answer

Yes, exactly when $a\ge b$ .

Full solution

The principal roots exist under the given conditions.
Their product simplifies to
$(\sqrt a+\sqrt b)(\sqrt a-\sqrt b)=a-b$

The difference $a-b$ is zero or positive precisely when $a\ge b$ .
Equality includes $a=b$ , including when both are zero.

Answer

Yes, exactly when $a\ge b$ .

Key idea

A conjugate product can make its sign clear through a difference of radicands.
Problem 9: Mark complete
Problem 10 A claimed rewrite

For $t>0$ , decide whether
$\frac{\sqrt t\sqrt[3]{t}}{\sqrt[6]{t}}=\sqrt[3]{t^2}$
holds for every allowed input. Justify your decision.
1. Hint 1
  The positive-base condition allows you to compare exponents.
2. Hint 2
  Add the two numerator exponents and subtract the denominator exponent.
Answer

Yes, for every $t>0$ .

Full solution

The denominator is positive, so division is valid.
The exponent on the left is
$\frac12+\frac13-\frac16=\frac23$

The right side is also $t^{2/3}$ .
Thus the two expressions agree throughout the stated domain.

Answer

Yes, for every $t>0$ .

Key idea

Converting different indices to rational exponents can verify a radical identity.
Problem 10: Mark complete

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