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Fractional Exponents and 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 three-quarters power

Evaluate $625^{3/4}$ without a calculator.
1. Hint 1
  The two numbers in a fractional exponent do different jobs: one names a root and the other names a power. Decide which does which here.
2. Hint 2
  The denominator $4$ is the index, so begin with the fourth root of $625$ , the number whose fourth power is $625$ . A fourth power is the square of a square, and $625$ is $25^2$ .
3. Hint 3
  Raise the fourth root to the power the numerator names. Taking the root first keeps every number in the work small.
Answer

$625^{3/4} = 125$ .

Full solution

In $625^{3/4}$ the denominator $4$ is the index of a root and the numerator $3$ is a power.
Taking the root first keeps the numbers small, so read the power as
$625^{3/4} = \left(\sqrt[4]{625}\right)^3.$

The fourth root of $625$ is the non-negative number whose fourth power is $625$ .
Since $5^2 = 25$ and $25^2 = 625$ , the fourth power of $5$ is $625$ .
So
$\sqrt[4]{625} = 5.$
The number $-5$ also has fourth power $625$ , but an even root names the non-negative one.

Now cube the root:
$\begin{aligned} \left(\sqrt[4]{625}\right)^3 &= 5^3 \\ &= 125. \end{aligned}$

Check the root by raising it to the index: $5^4 = 625$ .
Taking the power first gives the same value, but only by way of $625^3$ , a nine-digit number whose fourth root is much harder to find.

Answer

$625^{3/4} = 125$ .

Key idea

In $a^{m/n}$ the denominator is the index of the root and the numerator is the power; taking the root first kept every number here small.
Problem 1: Mark complete
Problem 2 Simplifying a cube root

Simplify $\sqrt[3]{192}$ , leaving no perfect-cube factor other than $1$ under the root.
1. Hint 1
  A root splits across a product, so look for a way to write $192$ as a product in which one factor comes out of the cube root cleanly.
2. Hint 2
  Test the perfect cubes $8$ , $27$ , $64$ and $125$ as factors of $192$ . The largest one that divides it does the job in a single step.
3. Hint 3
  If you pull out a smaller perfect cube first, check the radicand that is left: it may still hold another perfect-cube factor.
Answer

$\sqrt[3]{192} = 4\sqrt[3]{3}$ .

Full solution

The product rule for roots, $\sqrt[3]{ab} = \sqrt[3]{a}\,\sqrt[3]{b}$ , lets a perfect-cube factor come out of the root.
So look for the largest perfect cube that divides $192$ .

Among the perfect cubes up to $192$ , the number $8$ divides it, $27$ does not, $64$ does, and $125$ does not.
The largest is $64$ , since $192 = 64 \cdot 3$ and $64 = 4^3$ .

Split the root and evaluate the part that comes out:
$\begin{aligned} \sqrt[3]{192} &= \sqrt[3]{64 \cdot 3} \\ &= \sqrt[3]{64}\,\sqrt[3]{3} \\ &= 4\sqrt[3]{3}. \end{aligned}$

The radicand left is $3$ , which has no perfect-cube factor other than $1$ , so the root is fully simplified.

Pulling out $8$ first also works but takes two rounds.
It gives $2\sqrt[3]{24}$ , and since $24 = 8 \cdot 3$ still holds the perfect cube $8$ , a second round gives $2 \cdot 2\sqrt[3]{3}$ , the same $4\sqrt[3]{3}$ .

Check by cubing: $\left(4\sqrt[3]{3}\right)^3 = 4^3 \cdot 3$ , and
$64 \cdot 3 = 192.$

Answer

$\sqrt[3]{192} = 4\sqrt[3]{3}$ .

Key idea

Pulling out the largest perfect cube that divides the radicand simplifies a cube root in one step; a smaller one leaves a radicand to check again.
Problem 2: Mark complete
Problem 3 Adding and subtracting square roots

Simplify $\sqrt{54} - \sqrt{24} + \sqrt{96}$ as far as possible.
1. Hint 1
  The three radicands are different, so the terms are not like radicals as written. Simplify each root before trying to combine anything.
2. Hint 2
  Pull the largest perfect-square factor out of each radicand. All three radicands turn out to be a perfect square times the same number.
3. Hint 3
  Once every term is a whole number times the same radical, combine them as you would like terms, keeping each sign with its term.
Answer

$5\sqrt{6}$ .

Full solution

Unlike radicals do not combine, so first put each term in simplest form by pulling out its largest perfect-square factor.

Each radicand is $6$ times a perfect square: $54 = 9 \cdot 6$ , $24 = 4 \cdot 6$ and
$96 = 16 \cdot 6.$
Splitting each root with the product rule and taking the square root of the perfect square gives
$\begin{aligned} \sqrt{54} &= 3\sqrt{6}, \\ \sqrt{24} &= 2\sqrt{6}, \\ \sqrt{96} &= 4\sqrt{6}. \end{aligned}$

All three terms now carry the same radical $\sqrt{6}$ , so they are like radicals, and the distributive law combines them:
$\begin{aligned} 3\sqrt{6} - 2\sqrt{6} + 4\sqrt{6} &= (3 - 2 + 4)\sqrt{6} \\ &= 5\sqrt{6}. \end{aligned}$

The radicand $6$ has no perfect-square factor other than $1$ , so $5\sqrt{6}$ is fully simplified.

Answer

$5\sqrt{6}$ .

Key idea

Simplify every radical first, because terms that look unlike can turn out to share a radical and then combine as like terms.
Problem 3: Mark complete
Problem 4 Three roots as one power

For $x > 0$ , write $\dfrac{\sqrt[3]{x^{4}}\,\sqrt{x^{3}}}{\sqrt[6]{x}}$ as a single power of $x$ , with its exponent in lowest terms. Then write that power as a single radical.
1. Hint 1
  The rules for roots need a shared index, and these three roots have different ones. Powers of one base combine whatever their exponents, so work with powers of $x$ .
2. Hint 2
  Rewrite each root as a power of $x$ : the index becomes the denominator of the exponent and the power inside becomes the numerator.
3. Hint 3
  Add the exponents of the two factors on top and subtract the exponent on the bottom, over the common denominator $6$ . Reduce the result before turning it back into a radical, whose index is the denominator.
Answer

$x^{8/3}$ , and as a single radical $\sqrt[3]{x^{8}}$ . The same number, root first, is $\left(\sqrt[3]{x}\right)^{8}$ , and simplified, $x^{2}\sqrt[3]{x^{2}}$ , which is not itself a single radical.

Full solution

The product rule for roots needs a shared index, but these roots have indices $3$ , $2$ and $6$ .
Powers of the same base combine whatever their exponents, so rewrite each root as a power of $x$ , making the index the denominator and the power inside the numerator:
$\begin{aligned} \sqrt[3]{x^{4}} &= x^{4/3}, \\ \sqrt{x^{3}} &= x^{3/2}, \\ \sqrt[6]{x} &= x^{1/6}. \end{aligned}$

The two factors on top multiply, so the product rule adds their exponents.
The factor on the bottom divides, so the quotient rule subtracts its exponent.
The condition $x > 0$ keeps every root real and the bottom nonzero.
So the expression is
$x^{\frac{4}{3} + \frac{3}{2} - \frac{1}{6}}.$

Over the common denominator $6$ , the exponent is
$\begin{aligned} \frac{4}{3} + \frac{3}{2} - \frac{1}{6} &= \frac{8}{6} + \frac{9}{6} - \frac{1}{6} \\ &= \frac{16}{6} \\ &= \frac{8}{3}. \end{aligned}$
So the expression equals $x^{8/3}$ .

To turn $x^{8/3}$ into a radical, make the denominator $3$ the index and the numerator $8$ the power inside, which gives $\sqrt[3]{x^{8}}$ .
Read with the root first, the same number is $\left(\sqrt[3]{x}\right)^{8}$ .
Since $x^{8} = x^{6} \cdot x^{2}$ and $x^{6}$ is the perfect cube $\left(x^{2}\right)^{3}$ , pulling it out gives $x^{2}\sqrt[3]{x^{2}}$ , the same number again, though no longer a single radical.

Check with $x = 64$ , which is $2^6$ .
The three roots are then $2^8$ , $2^9$ and $2$ , so the expression is $2^{8 + 9 - 1}$ , which is $2^{16}$ .
The answer gives $64^{8/3} = 2^{16}$ as well.

Answer

$x^{8/3}$ , and as a single radical $\sqrt[3]{x^{8}}$ . The same number, root first, is $\left(\sqrt[3]{x}\right)^{8}$ , and simplified, $x^{2}\sqrt[3]{x^{2}}$ , which is not itself a single radical.

Key idea

Rewriting roots as fractional powers lets the exponent laws combine roots whose indices differ.
Problem 4: Mark complete
Problem 5 The faces of a cube-shaped box

A cube-shaped box has volume $V$ cubic centimeters. If each edge is $s$ centimeters long, then $V = s^3$ , and each of the six faces is a square of side $s$ .

Explain why the area of one face is $V^{2/3}$ square centimeters. Then find the total area of the six faces of a box whose volume is $512$ cubic centimeters.
1. Hint 1
  Go from the volume to a face in two stages: first find the edge from the volume, then find the area of a face from the edge.
2. Hint 2
  The edge is the number whose cube is $V$ , that is, its cube root. Write that cube root as a power of $V$ .
3. Hint 3
  A face's area is the edge squared, so square the power you just wrote and use the power-of-a-power rule. For a volume of $512$ , look for the whole number whose cube is $512$ .
Answer

Edge: $V^{1/3}$ centimeters; one face: its square, $V^{2/3}$ square centimeters. For $V = 512$ : a total area of $384$ square centimeters (edges of $8$ centimeters, faces of $64$ square centimeters).

Full solution

The edge $s$ satisfies $s^3 = V$ , so $s$ is the number whose cube is $V$ , the cube root of $V$ .
As a fractional exponent, the edge is
$s = V^{1/3}.$

A face is a square of side $s$ , so its area is $s^2$ .
Replace $s$ with $V^{1/3}$ and multiply the exponents by the power-of-a-power rule:
$\begin{aligned} s^2 &= \left(V^{1/3}\right)^2 \\ &= V^{2/3}. \end{aligned}$
So one face has area $V^{2/3}$ square centimeters.

The exponent records the two stages in the usual way.
Its denominator $3$ takes a cube root, which turns the volume into the edge, and its numerator $2$ squares, which turns the edge into the area of a face.

For $V = 512$ , take the cube root first.
Since $8^3 = 512$ , each edge is $8$ centimeters.
Then square it:
$\begin{aligned} 512^{2/3} &= \left(\sqrt[3]{512}\right)^2 \\ &= 8^2 \\ &= 64. \end{aligned}$
Each face has area $64$ square centimeters.

The six faces together have area $6 \cdot 64 = 384$ square centimeters.
As a check, a cube with edges of $8$ centimeters has volume $8^3 = 512$ cubic centimeters, as required.

Answer

Edge: $V^{1/3}$ centimeters; one face: its square, $V^{2/3}$ square centimeters. For $V = 512$ : a total area of $384$ square centimeters (edges of $8$ centimeters, faces of $64$ square centimeters).

Key idea

The exponent $\tfrac{2}{3}$ reads as two steps: a cube root that turns a cube's volume into its edge, and a square that turns the edge into the area of a face.
Problem 5: Mark complete
Problem 6 Two square plots on one road

Two square plots of land have areas of $44$ square meters and $176$ square meters. They sit next to each other along a straight road, each with one full side on the road and no gap between them.

Find the total length of road that the two plots touch, exactly and in simplest radical form. Sam adds the two areas first and says the length is $\sqrt{220}$ meters. Without a calculator, show that Sam's value is wrong.
1. Hint 1
  The side of a square is the square root of its area, and the road length is the sum of two sides. Work out each side before adding.
2. Hint 2
  Simplify $\sqrt{44}$ and $\sqrt{176}$ by pulling out perfect-square factors. If the results share a radical, they are like radicals and add.
3. Hint 3
  To test Sam's value, trap each side and $\sqrt{220}$ between whole numbers, using the perfect squares next to $44$ , $176$ and $220$ .
Answer

The road length is $6\sqrt{11}$ meters. Sam's value, $\sqrt{220}$ meters, is wrong and too short.

Full solution

A square's side is the square root of its area, so the plots have sides of $\sqrt{44}$ meters and $\sqrt{176}$ meters.
The road length is their sum, $\sqrt{44} + \sqrt{176}$ meters.

Simplify each side by pulling out its largest perfect-square factor, using $44 = 4 \cdot 11$ and $176 = 16 \cdot 11$ :
$\begin{aligned} \sqrt{44} &= 2\sqrt{11}, \\ \sqrt{176} &= 4\sqrt{11}. \end{aligned}$

Both sides now carry $\sqrt{11}$ , so they are like radicals and add by the distributive law:
$\begin{aligned} 2\sqrt{11} + 4\sqrt{11} &= (2 + 4)\sqrt{11} \\ &= 6\sqrt{11}. \end{aligned}$
The road length is $6\sqrt{11}$ meters.

Now test Sam's value by size.
Since $36 < 44$ , the side $\sqrt{44}$ is more than $6$ , and since $169 < 176$ , the side $\sqrt{176}$ is more than $13$ .
So the road length is more than $19$ meters.
But $220 < 225$ , so $\sqrt{220}$ is less than $15$ .
Sam's value is too small to be the road length.

Sam turned a sum of two roots into the root of a sum, and no rule allows that: a root splits across a product, not across a sum.
Written as a single root, the correct length is $\sqrt{36}\,\sqrt{11}$ , which the product rule makes $\sqrt{396}$ meters, not $\sqrt{220}$ .

Answer

The road length is $6\sqrt{11}$ meters. Sam's value, $\sqrt{220}$ meters, is wrong and too short.

Key idea

Simplify square roots into like radicals before adding them; adding the numbers under the roots first, as Sam did, gives a different number.
Problem 6: Mark complete
Problem 7 Kai's negative exponent

Kai evaluates $\left(\tfrac{1000}{343}\right)^{-2/3}$ and writes
$\begin{aligned} \left(\tfrac{1000}{343}\right)^{-2/3} &= -\left(\tfrac{1000}{343}\right)^{2/3} \\ &= -\left(\frac{\sqrt[3]{1000}}{\sqrt[3]{343}}\right)^{2} \\ &= -\left(\tfrac{10}{7}\right)^{2} \\ &= -\tfrac{100}{49}. \end{aligned}$
Find the first step where Kai goes wrong, and explain the mistake. Then evaluate $\left(\tfrac{1000}{343}\right)^{-2/3}$ correctly, and explain how the sign of Kai's answer alone shows that it cannot be right.
1. Hint 1
  Ask what the minus sign in an exponent tells you to do. Does it make the value negative, or does it call for something else?
2. Hint 2
  A negative exponent means a reciprocal: $b^{-p}$ is $\tfrac{1}{b^{p}}$ whenever $b^{p}$ is defined and nonzero, as it is for this positive base. Check each of Kai's steps against that.
3. Hint 3
  Kai's work on the positive power $\tfrac{2}{3}$ can be reused: once you have that value, take its reciprocal. For the last part, think about the sign of any power of a positive number.
Answer

Kai's first step is wrong: it treats the negative exponent as a minus sign instead of a reciprocal. The correct value is $\tfrac{49}{100}$ (or $0.49$ ). Kai's answer has the wrong sign.

Full solution

Kai's first step replaces the exponent $-\tfrac{2}{3}$ with a minus sign in front of the power, and that is the mistake.
A negative exponent asks for a reciprocal: whenever $b^{p}$ is defined and nonzero, the power $b^{-p}$ is $\tfrac{1}{b^{p}}$ , not $-b^{p}$ .

His later steps are correct moves, but they act on the wrong expression.
The cube root of the top and of the bottom gives $\tfrac{10}{7}$ , and squaring gives $\tfrac{100}{49}$ , so his stray minus sign carries through to $-\tfrac{100}{49}$ .

Correctly, the negative exponent asks for the reciprocal of the positive power:
$\left(\tfrac{1000}{343}\right)^{-2/3} = \frac{1}{\left(\frac{1000}{343}\right)^{2/3}}.$

Kai's own later steps show that the positive power is $\tfrac{100}{49}$ .
The denominator $3$ takes the cube root of the top and the bottom, and since $10^3 = 1000$ and $7^3 = 343$ , that gives $\tfrac{10}{7}$ ; the numerator $2$ then squares it.
So
$\begin{aligned} \left(\tfrac{1000}{343}\right)^{-2/3} &= 1 \div \tfrac{100}{49} \\ &= \tfrac{49}{100}. \end{aligned}$
As a decimal, that is $0.49$ .

Flipping the base first gives the same value.
The reciprocal of $\tfrac{1000}{343}$ is $\tfrac{343}{1000}$ , whose cube root is $\tfrac{7}{10}$ , and squaring $\tfrac{7}{10}$ gives $\tfrac{49}{100}$ again.

The sign alone gives Kai's answer away.
The base $\tfrac{1000}{343}$ is positive, its cube root is positive, a power of a positive number is positive, and so is the reciprocal of a positive number.
So the value must be positive, and a negative answer cannot be right.

Answer

Kai's first step is wrong: it treats the negative exponent as a minus sign instead of a reciprocal. The correct value is $\tfrac{49}{100}$ (or $0.49$ ). Kai's answer has the wrong sign.

Key idea

A negative exponent calls for a reciprocal, not a negative value, and a positive base raised to a fractional power stays positive.
Problem 7: Mark complete
Problem 8 Which square root is $121^{1/2}$ ?

Jess notices that $11^2 = 121$ and $(-11)^2 = 121$ . She says that $121^{1/2}$ could just as well be $-11$ , which would make $121^{3/2}$ equal to $-1331$ .

Explain, using the product rule for exponents, why $121^{1/2}$ must square to $121$ , and what then decides between $11$ and $-11$ . Then find the value of $121^{3/2}$ .
1. Hint 1
  Two questions are separate here: what the exponent laws force $121^{1/2}$ to satisfy, whatever it means, and what is still left to choose once they have.
2. Hint 2
  Apply the product rule, adding the exponents, to $121^{1/2} \cdot 121^{1/2}$ . Then ask which numbers square to the result, and whether that equation prefers either one.
3. Hint 3
  A symbol has to name a single number, and the radical sign already chose one square root. Once $121^{1/2}$ is fixed, read $121^{3/2}$ as a square root followed by a cube.
Answer

The product rule makes $121^{1/2} \cdot 121^{1/2} = 121$ . The principal-root convention decides between $11$ and $-11$ : $121^{1/2} = 11$ . The value is $121^{3/2} = 1331$ , not $-1331$ .

Full solution

Apply the product rule, adding the exponents, to $121^{1/2}$ times itself:
$\begin{aligned} 121^{1/2} \cdot 121^{1/2} &= 121^{\frac{1}{2} + \frac{1}{2}} \\ &= 121^{1} \\ &= 121. \end{aligned}$
So whatever $121^{1/2}$ means, it is a number whose square is $121$ , a square root of $121$ .

That equation alone leaves two candidates, since $11 \cdot 11 = 121$ and
$(-11)(-11) = 121.$
Jess is right that $-11$ passes this test: the equation asks only that the number square to $121$ , so it does not choose between them.

What decides is a convention, the same one the radical sign already follows.
A symbol has to name one number, so $a^{1/2}$ is defined to be the non-negative square root.
So $121^{1/2}$ is $\sqrt{121}$ , which is $11$ .
With Jess's choice the two notations would disagree, since $121^{1/2}$ would be $-11$ while $\sqrt{121}$ is $11$ .

With the root fixed, read $121^{3/2}$ with the denominator as a square root and the numerator as a cube:
$\begin{aligned} 121^{3/2} &= \left(\sqrt{121}\right)^3 \\ &= 11^3 \\ &= 1331. \end{aligned}$
Jess's $-1331$ comes from cubing the negative root, which the convention rules out.

Answer

The product rule makes $121^{1/2} \cdot 121^{1/2} = 121$ . The principal-root convention decides between $11$ and $-11$ : $121^{1/2} = 11$ . The value is $121^{3/2} = 1331$ , not $-1331$ .

Key idea

The product rule makes $a^{1/2}$ a square root of $a$ , and for $a \ge 0$ the principal-root convention picks the non-negative one, $\sqrt{a}$ .
Problem 8: Mark complete
Problem 9 Powers of $-216$

Evaluate $(-216)^{2/3}$ and $(-216)^{-1/3}$ . Then decide whether $(-216)^{3/2}$ is a real number, and explain what makes its case different from that of $(-216)^{2/3}$ .
1. Hint 1
  With a negative base, look at the denominator of the exponent first. It names a root, and whether that root exists depends on whether its index is even or odd.
2. Hint 2
  For $(-216)^{2/3}$ , take the cube root first: which number cubed gives $-216$ ? Then square it. For $(-216)^{-1/3}$ , take the reciprocal of that cube root.
3. Hint 3
  In $(-216)^{3/2}$ the denominator asks for a square root of $-216$ . Ask whether any real number squares to a negative, and whether taking the cube first would help.
Answer

$(-216)^{2/3} = 36$ and $(-216)^{-1/3} = -\tfrac{1}{6}$ . $(-216)^{3/2}$ is not a real number. The difference is the denominator: $2$ in $(-216)^{3/2}$ is even, and $3$ in $(-216)^{2/3}$ is odd.

Full solution

In $(-216)^{2/3}$ the denominator $3$ is odd, and an odd root accepts a negative radicand, because an odd number of negative factors has a negative product.
Since $(-6)^3 = -216$ , the cube root is
$\sqrt[3]{-216} = -6.$

The numerator $2$ then squares that root:
$\begin{aligned} (-216)^{2/3} &= \left(\sqrt[3]{-216}\right)^2 \\ &= (-6)^2 \\ &= 36. \end{aligned}$
Squaring removes the negative sign, so this power is positive.

In $(-216)^{-1/3}$ the minus sign calls for a reciprocal and the $\tfrac{1}{3}$ for a cube root.
The denominator is odd and the base is not zero, so the value exists:
$\begin{aligned} (-216)^{-1/3} &= \frac{1}{\sqrt[3]{-216}} \\ &= \frac{1}{-6} \\ &= -\frac{1}{6}. \end{aligned}$
A reciprocal keeps the sign of the number it flips, so this power is negative.

In $(-216)^{3/2}$ the denominator $2$ asks for $\sqrt{-216}$ before any cube.
An even root needs a radicand that is zero or positive, because the square of every real number is zero or positive.
No real number squares to $-216$ , so this route gives no real value.

Taking the cube first does not rescue it.
The cube $(-216)^3$ is negative, so $\sqrt{(-216)^3}$ is again a square root of a negative number.
So $(-216)^{3/2}$ is not a real number, while $(-216)^{2/3}$ is $36$ .
Swapping the numerator and the denominator changed the root from an odd one, which a negative number has, to an even one, which it does not.

Answer

$(-216)^{2/3} = 36$ and $(-216)^{-1/3} = -\tfrac{1}{6}$ . $(-216)^{3/2}$ is not a real number. The difference is the denominator: $2$ in $(-216)^{3/2}$ is even, and $3$ in $(-216)^{2/3}$ is odd.

Key idea

For a negative base, with the exponent in lowest terms, an odd denominator gives a real value and an even denominator gives none.
Problem 9: Mark complete
Problem 10 True for every positive number?

Decide whether each statement below is true for every positive number $a$ . Prove each true one, and show that each false one fails for a particular positive value of $a$ .

Statement A: $\sqrt{a} + \sqrt{a} = \sqrt{4a}$

Statement B: $\sqrt{a} + \sqrt[3]{a} = 2\sqrt{a}$
1. Hint 1
  A statement that is true for every positive $a$ needs an argument that works for all of them at once; a false one needs just one value where the two sides differ.
2. Hint 2
  Both statements add two roots. Two roots combine into one term when they are like radicals, with the same index and the same radicand. For statement A, split the right side $\sqrt{4a}$ with the product rule for roots.
3. Hint 3
  To test a value, choose one where every root in the statement is a whole number, such as a perfect sixth power. A value where the two sides happen to agree proves nothing about the others.
Answer

Statement A is true. Statement B is false; for example, at $a = 64$ the left side is $12$ and the right side is $16$ .

Full solution

A true statement needs an argument that covers every positive $a$ at once, and a false one needs only one positive value of $a$ where the two sides differ.

Statement A is true.
The two terms are like radicals, so the distributive law combines them, and the product rule for roots then brings the $2$ under the root as $\sqrt{4}$ :
$\begin{aligned} \sqrt{a} + \sqrt{a} &= 2\sqrt{a} \\ &= \sqrt{4}\,\sqrt{a} \\ &= \sqrt{4a}. \end{aligned}$
Adding the radicands instead would give $\sqrt{2a}$ , which is a different number.

Statement B is false.
The terms $\sqrt{a}$ and $\sqrt[3]{a}$ share a radicand but not an index, so they are not like radicals and do not combine into $2\sqrt{a}$ .
Take $a = 64$ , whose square root and cube root are whole numbers.
The left side is $8 + 4$ , which is $12$ , and the right side is $2 \cdot 8$ , which is $16$ .

At $a = 1$ the two sides of statement B agree, since both equal $2$ .
One agreeing value proves nothing: the statement fails at $a = 64$ , so it is not true for every positive $a$ .

Answer

Statement A is true. Statement B is false; for example, at $a = 64$ the left side is $12$ and the right side is $16$ .

Key idea

Like radicals, with the same index and the same radicand, add by the distributive law, so $\sqrt{a} + \sqrt{a}$ is $2\sqrt{a}$ , which is $\sqrt{4a}$ ; a shared radicand with different indices, as in $\sqrt{a} + \sqrt[3]{a}$ , does not make two roots like radicals.
Problem 10: Mark complete

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