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Rationalizing Denominators: 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 single root in the way

Rationalize and simplify $\frac4{\sqrt{14}}$ , so that no radical is left in the denominator.
1. Hint 1
  Multiplying a fraction by a nonzero quantity divided by itself changes its form but not its value.
2. Hint 2
  A square root multiplied by itself gives back the number under it, so use $\sqrt{14}$ on both the numerator and the denominator, then look for a common factor.
Answer

$\frac{2\sqrt{14}}7$ .

Full solution

Multiply the numerator and the denominator by the nonzero number $\sqrt{14}$ , which is multiplication by one.
The denominator becomes $14$ and the numerator becomes $4\sqrt{14}$ , so
$\frac4{\sqrt{14}}=\frac{4\sqrt{14}}{14}.$
The numbers $4$ and $14$ share a factor of $2$ , so the fraction reduces to $\frac{2\sqrt{14}}7$ .
As a check, $4/\sqrt{14}\approx1.069$ and $2\sqrt{14}/7\approx1.069$ .

Answer

$\frac{2\sqrt{14}}7$ .

Key idea

Multiplying by a root over itself clears a single root from a denominator, and the result may still reduce.
Problem 1: Mark complete
Problem 2 A denominator change

Find the numerator $N$ that makes $\frac{7}{\sqrt{11}}=\frac{N}{11}$ .
1. Hint 1
  An equivalent fraction requires the same nonzero multiplier on its numerator and denominator.
2. Hint 2
  Determine which multiplier changes $\sqrt{11}$ into $11$ .
Answer

$N=7\sqrt{11}$ .

Full solution

The needed multiplier is $\sqrt{11}$ , which is nonzero.
Multiplying both parts by it gives
$\frac{7}{\sqrt{11}}=\frac{7\sqrt{11}}{11}.$
Thus $N=7\sqrt{11}$ , and dividing numerator and denominator by $\sqrt{11}$ returns the original fraction.

Answer

$N=7\sqrt{11}$ .

Key idea

Changing a denominator while preserving a fraction requires applying the same multiplier to its numerator.
Problem 2: Mark complete
Problem 3 A denominator with two roots

Write $\frac2{\sqrt{10}-\sqrt7}$ in simplest exact form with no radical in its denominator.
1. Hint 1
  Multiplying by just one of these roots still leaves a root in the denominator, because $\sqrt{10}$ and $\sqrt7$ do not simplify to the same root.
2. Hint 2
  Pair the denominator with the same two roots joined by the opposite sign. Their product is a difference of squares, and the numerator must be multiplied by that same expression.
Answer

$\frac{2(\sqrt{10}+\sqrt7)}3$ , or equivalently $\frac{2\sqrt{10}+2\sqrt7}3$ .

Full solution

The conjugate of $\sqrt{10}-\sqrt7$ is $\sqrt{10}+\sqrt7$ , which is nonzero because both roots are positive.
Multiplying the numerator and the denominator by it makes the denominator a difference of squares,
$10-7=3.$
The numerator becomes $2(\sqrt{10}+\sqrt7)$ , whose coefficient $2$ shares no factor above $1$ with $3$ , so the fraction does not reduce, and the result is $\frac{2(\sqrt{10}+\sqrt7)}3$ .
As a check, $2/(\sqrt{10}-\sqrt7)\approx3.872$ and $2(\sqrt{10}+\sqrt7)/3\approx3.872$ .

Answer

$\frac{2(\sqrt{10}+\sqrt7)}3$ , or equivalently $\frac{2\sqrt{10}+2\sqrt7}3$ .

Key idea

The conjugate turns a two-root denominator into the plain difference of the two radicands.
Problem 3: Mark complete
Problem 4 Combined measurements

Two lengths are $\frac4{\sqrt{20}}$ cm and $\frac3{\sqrt{45}}$ cm. Find their total length in simplest exact form with no radical in a denominator.
1. Hint 1
  Simplify each radical before adding the lengths.
2. Hint 2
  Rationalize the reduced fractions so their matching radical terms can be combined.
Answer

$\frac{3\sqrt5}5$ cm.

Full solution

The radicals simplify to $2\sqrt5$ and $3\sqrt5$ .
The lengths become $2/\sqrt5$ cm and $1/\sqrt5$ cm.
Multiplying both parts of each fraction by the nonzero number $\sqrt5$ gives $2\sqrt5/5$ cm and $\sqrt5/5$ cm.
The two rationalized lengths add to
$\frac{2\sqrt5}5+\frac{\sqrt5}5=\frac{3\sqrt5}5.$
The total length is therefore $\frac{3\sqrt5}5$ cm.
Multiplying that total by $\sqrt5$ gives $3$ , and multiplying the unrationalized sum $3/\sqrt5$ by $\sqrt5$ gives $3$ as well, so the two forms agree.

Answer

$\frac{3\sqrt5}5$ cm.

Key idea

Simplifying each radical first turns two unlike fractions into like terms that can be added.
Problem 4: Mark complete
Problem 5 A rectangle from its area

A rectangle has area $15$ square cm and width $\sqrt{11}-\sqrt5$ cm. Find its perimeter in simplest exact form with no radical in a denominator.
1. Hint 1
  Divide the area by the positive width to find the other dimension.
2. Hint 2
  Use the conjugate to simplify that quotient, then add the dimensions and double their sum.
Answer

$7\sqrt{11}+3\sqrt5$ cm.

Full solution

The width is positive because $11>5$ .
Its conjugate product is
$11-5=6.$
Multiply the numerator and the denominator of $15/(\sqrt{11}-\sqrt5)$ by $\sqrt{11}+\sqrt5$ .
The numerator becomes $15(\sqrt{11}+\sqrt5)$ , and dividing by the denominator $6$ gives a length of $\frac{5(\sqrt{11}+\sqrt5)}2$ cm.
Twice that length is $5\sqrt{11}+5\sqrt5$ cm and twice the width is $2\sqrt{11}-2\sqrt5$ cm, so the perimeter is $7\sqrt{11}+3\sqrt5$ cm.
Checking the area, the length times the width is $\frac{5\cdot6}2=15$ square cm.

Answer

$7\sqrt{11}+3\sqrt5$ cm.

Key idea

Rationalizing an area-to-width quotient can produce an exact perimeter with simplified radical terms.
Problem 5: Mark complete
Problem 6 Three values of $b$

For each of $b=3$ , $b=6$ and $b=13$ , let $T=\frac1{3+\sqrt b}+\frac1{3-\sqrt b}$ . Find $T$ for each of the three, and state which of them make $T$ an integer.
1. Hint 1
  The two denominators are conjugates, and neither of them is zero for any of the three listed values.
2. Hint 2
  Add the fractions over their product, then substitute each value of $b$ into the combined form.
Answer

$b=3$ gives $T=1$ and $b=6$ gives $T=2$ , both integers; $b=13$ gives $T=-\frac32$ , which is not an integer.

Full solution

Adding the fractions over their common denominator, the numerators $3-\sqrt b$ and $3+\sqrt b$ add to $6$ , and the denominator product is $9-b$ .
Therefore
$T=\frac6{9-b}.$
The three values give $6/6=1$ , $6/3=2$ and $6/(-4)=-\frac32$ , so the first two are integers and the third is not.

Since $\sqrt{13}>3$ , the denominator $3-\sqrt{13}$ is negative rather than zero, which is why $T$ is negative there.
None of the three values is $9$ , so no denominator, original or combined, is zero.

Answer

$b=3$ gives $T=1$ and $b=6$ gives $T=2$ , both integers; $b=13$ gives $T=-\frac32$ , which is not an integer.

Key idea

Combining fractions with conjugate denominators can cancel every radical before any value is substituted.
Problem 6: Mark complete
Problem 7 One less than $r$

A number $r$ is given by $r=\frac{\sqrt7+2}{\sqrt7-2}$ . Find $r-1$ in simplest exact form with a rational denominator.
1. Hint 1
  Subtracting one can simplify the numerator before rationalizing.
2. Hint 2
  Write one with the same denominator, then multiply the resulting numerator and denominator by the conjugate.
Answer

$\frac{8+4\sqrt7}3$ , or equivalently $\frac{4(\sqrt7+2)}3$ .

Full solution

The denominator is nonzero because $\sqrt7>2$ .
Subtracting one gives
$r-1=\frac4{\sqrt7-2}.$
Multiplying both parts by the nonzero conjugate $\sqrt7+2$ gives
$r-1=\frac{4(\sqrt7+2)}3.$
Expanding the numerator yields $(8+4\sqrt7)/3$ , which is about $6.194$ .
Adding one back gives $(11+4\sqrt7)/3$ , the rationalized value of $r$ .

Answer

$\frac{8+4\sqrt7}3$ , or equivalently $\frac{4(\sqrt7+2)}3$ .

Key idea

Subtracting a simple quantity before rationalizing can reduce the work in the numerator.
Problem 7: Mark complete
Problem 8 A formula range

A student claims that $\frac1{\sqrt n+2}=\frac{\sqrt n-2}{n-4}$ for every positive integer $n$ . Decide whether the claim is correct, and state exactly which positive integers allow the equality. Explain what happens at any excluded input.
1. Hint 1
  An equivalent fraction must have a defined value on both sides.
2. Hint 2
  The product of $\sqrt n+2$ and $\sqrt n-2$ is $n-4$ . Check when the second factor can be zero.
3. Hint 3
  At that input, compare the original denominator with the proposed new denominator. For all other inputs, consider multiplying by the conjugate divided by itself.
Answer

The claim is false. The equality holds for every positive integer $n\ne4$ . At $n=4$ , the left side is $\frac14$ and the right side is undefined.

Full solution

The original denominator $\sqrt n+2$ is positive for every positive integer $n$ , so the original fraction is always defined.

The conjugate product is
$(\sqrt n+2)(\sqrt n-2)=n-4.$
The conjugate is zero exactly when $n=4$ .
At that input, the original fraction is $\frac14$ , while the proposed fraction has numerator and denominator both zero and is undefined.

If $n\ne4$ , the conjugate is nonzero.
Multiplying numerator and denominator by it is multiplication by one, so
$\frac1{\sqrt n+2}=\frac{\sqrt n-2}{n-4}.$
Thus the equality holds for every positive integer except $4$ .

Answer

The claim is false. The equality holds for every positive integer $n\ne4$ . At $n=4$ , the left side is $\frac14$ and the right side is undefined.

Key idea

A conjugate gives an equivalent fraction only when the conjugate used as a multiplier is nonzero.
Problem 8: Mark complete
Problem 9 Neighboring radical values

For every positive integer $n$ , Ada claims that $\frac1{\sqrt{n+1}+\sqrt n}=\sqrt{n+1}-\sqrt n$ . Decide whether the claim is correct, including when one of the square roots is an integer.
1. Hint 1
  Check the product of the two expressions made by adding and subtracting the square roots.
2. Hint 2
  The two radicands differ by one, and neither the sum nor the difference of the roots is zero.
Answer

The claim is correct for every positive integer $n$ .

Full solution

The conjugate product is
$(n+1)-n=1.$
Since $n+1>n>0$ , the sum and difference of the two square roots are both nonzero.
Multiplying the proposed right side by the original denominator gives $1$ , so it is exactly the reciprocal.
Nothing in the product $(n+1)-n=1$ depended on whether a root is irrational, so the argument still holds when one of the roots is an integer.
At $n=15$ , for example, the claim reads $\frac1{4+\sqrt{15}}=4-\sqrt{15}$ , and the conjugate product is still $16-15=1$ .

Answer

The claim is correct for every positive integer $n$ .

Key idea

Conjugate radical expressions whose squared terms differ by one are reciprocals.
Problem 9: Mark complete
Problem 10 A claim about the only method

A student says that $\frac1{\sqrt{300}-\sqrt{75}}$ can be rationalized only by using a conjugate. Rationalize the expression and decide whether the student is right.
1. Hint 1
  Look at the two radicands before choosing how to start: each one has a perfect-square factor.
2. Hint 2
  One route starts by simplifying each radical; the other multiplies the original fraction by its conjugate. Carry out both and compare what they give.
Answer

$\frac{\sqrt3}{15}$ . The student is wrong.

Full solution

Simplifying first, $\sqrt{300}=10\sqrt3$ and $\sqrt{75}=5\sqrt3$ , so the denominator is the single root $5\sqrt3$ .
Multiplying both parts by $\sqrt3$ makes the denominator $5\cdot3=15$ and the numerator $\sqrt3$ , so
$\frac1{5\sqrt3}=\frac{\sqrt3}{15}.$
That route used no conjugate, so the student's claim is false.

The conjugate route works too.
Multiply both parts of the original fraction by the nonzero conjugate $\sqrt{300}+\sqrt{75}$ .
The numerator becomes $10\sqrt3+5\sqrt3=15\sqrt3$ and the denominator becomes $300-75=225$ , so
$\frac{15\sqrt3}{225}=\frac{\sqrt3}{15}.$
The two routes agree, which checks the work.

Answer

$\frac{\sqrt3}{15}$ . The student is wrong.

Key idea

Simplifying like radicals may leave a single-root denominator, so a conjugate is not the only way to clear a two-term denominator.
Problem 10: Mark complete

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