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Rationalizing and Radical Conjugates: 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 denominator entry

A nonzero rational number $D$ satisfies
$\frac1{\sqrt6-1}=\frac{\sqrt6+1}{D}$
Find $D$ .
1. Hint 1
  The matching sum and difference give a difference of squares.
2. Hint 2
  Multiply the original denominator by the numerator in the proposed form.
Answer

$D=5$ .

Full solution

The original denominator is nonzero since $\sqrt6>1$ .
Multiplying through gives
$D=(\sqrt6-1)(\sqrt6+1)$

The product is $6-1$ , so $D=5$ .
Substituting this denominator gives the original reciprocal.

Answer

$D=5$ .

Key idea

A conjugate product determines the rational denominator in an equivalent fraction.
Problem 1: Mark complete
Problem 2 A fourth root below

Write $3/\sqrt[4]{1331}$ with a rational denominator and in simplest radical form.
1. Hint 1
  Write $1331$ as a power of $11$ .
2. Hint 2
  The denominator has exponent $3/4$ , so multiply by the missing quarter power.
Answer

$\frac{3\sqrt[4]{11}}{11}$ .

Full solution

The positive denominator is $11^{3/4}$ .
Multiply numerator and denominator by $11^{1/4}$ .
The denominator becomes $11$ , so
$\frac3{\sqrt[4]{1331}}=\frac{3\sqrt[4]{11}}{11}$

Multiplying the result by the original denominator gives $\frac{3\cdot11}{11}=3$ , confirming the quotient.

Answer

$\frac{3\sqrt[4]{11}}{11}$ .

Key idea

Completing the missing power can clear a higher root from a denominator.
Problem 2: Mark complete
Problem 3 A reduced quotient

Write $\dfrac{8\sqrt{41}}{\sqrt{41}(\sqrt{41}+3)}$ in the form $p+q\sqrt{41}$ , with rational $p,q$ .
1. Hint 1
  A nonzero common factor can be canceled before the denominator is changed.
2. Hint 2
  The remaining denominator pairs with $\sqrt{41}-3$ .
Answer

$-\frac34+\frac14\sqrt{41}$ .

Full solution

Both denominator factors are positive.
Cancel the common factor $\sqrt{41}$ to obtain $8/(\sqrt{41}+3)$ .
The conjugate product is $41-9=32$ , so
$\frac8{\sqrt{41}+3}=\frac{8(\sqrt{41}-3)}{32}$

That is $-\frac34+\frac14\sqrt{41}$ .
Multiplying by $\sqrt{41}+3$ returns $\frac{41-9}4=8$ , verifying the reduced quotient.

Answer

$-\frac34+\frac14\sqrt{41}$ .

Key idea

Simplifying common factors first can shorten a radical division.
Problem 3: Mark complete
Problem 4 A rectangle's proportions

A rectangle has perimeter $20$ cm and width $\sqrt{43}-1$ cm. Find its length, then find the ratio of its length to its width, written with a rational denominator.
1. Hint 1
  Half the perimeter is the sum of the length and the width.
2. Hint 2
  For the ratio, multiply numerator and denominator by the partner of $\sqrt{43}-1$ .
Answer

Length $11-\sqrt{43}$ cm; ratio $\frac{5\sqrt{43}-16}{21}$ .

Full solution

Half the perimeter is $10$ , so the length is
$10-(\sqrt{43}-1)=11-\sqrt{43}$
It is positive because $43<121$ , and the width is positive because $43>1$ .

The ratio is $\frac{11-\sqrt{43}}{\sqrt{43}-1}$ .
Multiplying by $\sqrt{43}+1$ over itself, the denominator becomes $43-1=42$ and the numerator becomes
$\begin{aligned}&(11-\sqrt{43})(\sqrt{43}+1)\\&=10\sqrt{43}-32\end{aligned}$

So the ratio is $\frac{10\sqrt{43}-32}{42}$ , which reduces to $\frac{5\sqrt{43}-16}{21}$ .

Answer

Length $11-\sqrt{43}$ cm; ratio $\frac{5\sqrt{43}-16}{21}$ .

Key idea

Rationalizing a ratio of exact lengths gives a form whose denominator is rational.
Problem 4: Mark complete
Problem 5 Two rational coefficients

Write $\dfrac{2+\sqrt{47}}{7+\sqrt{47}}$ as $p+q\sqrt{47}$ and identify the rational numbers $p$ and $q$ . Explain why the new denominator is nonzero.
1. Hint 1
  Look for a product whose radical terms cancel.
2. Hint 2
  Multiply numerator and denominator by $7-\sqrt{47}$ , and compute the new denominator before expanding the numerator.
Answer

$\frac{-33+5\sqrt{47}}2$ ; $p=-\frac{33}2$ , $q=\frac52$ ; the new denominator is $2$ .

Full solution

The original denominator is positive.
The conjugate product is
$49-47=2$
which is not zero.

Expanding the numerator $(2+\sqrt{47})(7-\sqrt{47})$ gives the rational term $14-47$ and the radical term $(-2+7)\sqrt{47}$ .
The quotient is
$\frac{-33+5\sqrt{47}}2$

Both coefficients are rational.
Multiplying the result by $7+\sqrt{47}$ recovers $2+\sqrt{47}$ .

Answer

$\frac{-33+5\sqrt{47}}2$ ; $p=-\frac{33}2$ , $q=\frac52$ ; the new denominator is $2$ .

Key idea

Dividing two numbers built from the same square root can return another number in the same form.
Problem 5: Mark complete
Problem 6 A cube root quotient

Write $\dfrac{\sqrt[3]{11}+\sqrt[3]{121}}{2+\sqrt[3]{11}}$ with a rational denominator and a numerator in simplest radical form.
1. Hint 1
  A cube-root binomial needs the multiplier from a sum of cubes.
2. Hint 2
  Let $t=\sqrt[3]{11}$ and multiply by $4-2t+t^2$ ; use $t^3=11$ to simplify.
Answer

$\frac{-11+15\sqrt[3]{11}+2\sqrt[3]{121}}{19}$ .

Full solution

Let $t=\sqrt[3]{11}$ , so $t^2=\sqrt[3]{121}$ and $t^3=11$ .
The denominator is positive.
Its three-term partner gives
$(2+t)(4-2t+t^2)=19$

Multiplying the numerator $t+t^2$ by that partner gives $4t+2t^2-t^3+t^4$ .
Since $t^4=11t$ , the simplified numerator is
$-11+15t+2t^2$

Dividing by the nonzero rational number $19$ gives the stated result.
Multiplying its numerator by $2+t$ yields $19t+19t^2$ , confirming the quotient.

Answer

$\frac{-11+15\sqrt[3]{11}+2\sqrt[3]{121}}{19}$ .

Key idea

A cube identity clears a cube-root binomial while its power relation simplifies the remaining numerator.
Problem 6: Mark complete
Problem 7 A difference quotient

For $x>0$ , rewrite $\dfrac{\sqrt{x+9}-3}{x}$ as one fraction whose numerator is $1$ , and keep its domain restriction.
1. Hint 1
  A matching sum changes the numerator to a difference of squares.
2. Hint 2
  The resulting numerator has a factor $x$ , which is nonzero on the stated domain.
Answer

$\frac1{\sqrt{x+9}+3}$ , with $x>0$ .

Full solution

The factor $\sqrt{x+9}+3$ is positive, so multiplying by it over itself preserves the value.
The new numerator is $x+9-9$ , namely $x$ .

Cancel the nonzero $x$ to obtain
$\frac1{\sqrt{x+9}+3}$

The rewritten expression is used with the original restriction $x>0$ , even though its displayed formula has a larger natural domain.

Answer

$\frac1{\sqrt{x+9}+3}$ , with $x>0$ .

Key idea

Changing a radical numerator can expose a canceling factor while the original domain remains in force.
Problem 7: Mark complete
Problem 8 Two kinds of conjugate

For $r=2+\sqrt{31}$ , find its complex conjugate and its radical conjugate. Compute the product of $r$ with each, and explain which product is rational.
1. Hint 1
  Decide whether $r$ has an imaginary part.
2. Hint 2
  The radical conjugate changes the sign of the $\sqrt{31}$ term.
Answer

Complex conjugate $2+\sqrt{31}$ , product $35+4\sqrt{31}$ ; radical conjugate $2-\sqrt{31}$ , product $-27$ . Only the product with the radical conjugate is rational.

Full solution

The number $r$ is real, so its complex conjugate is $r$ itself, and
$r\cdot r=35+4\sqrt{31}$
since
$4+4\sqrt{31}+31=35+4\sqrt{31}.$

The radical conjugate changes the sign of the radical term.
Since $4-31=-27$ ,
$(2+\sqrt{31})(2-\sqrt{31})=-27$

Only the radical conjugate cancels the $\sqrt{31}$ terms, because the product of the two radical terms is $-31$ , a rational number.
For this $r$ , the complex-conjugate product keeps the irrational term $4\sqrt{31}$ .

Answer

Complex conjugate $2+\sqrt{31}$ , product $35+4\sqrt{31}$ ; radical conjugate $2-\sqrt{31}$ , product $-27$ . Only the product with the radical conjugate is rational.

Key idea

For rational $a$ and $b$ and a positive integer $d$ , the radical conjugate makes $(a+b\sqrt d)(a-b\sqrt d)$ rational, while the complex conjugate of a real number is the number itself.
Problem 8: Mark complete
Problem 9 A claimed guarantee

A student says that whenever $a+b\sqrt d$ is nonzero, its partner $a-b\sqrt d$ must also be nonzero for rational $a,b$ and a positive integer $d$ . Test this claim with $a=3$ , $b=1$ , $d=9$ , and identify a missing condition that would make the guarantee valid.
1. Hint 1
  Evaluate both expressions at the given values first.
2. Hint 2
  If the partner were zero and $b\ne0$ , what would $a/b$ equal?
Answer

The claim fails: $a+b\sqrt d=6$ , $a-b\sqrt d=0$ ; require $d$ to be a positive nonsquare integer.

Full solution

With the supplied values, $\sqrt9=3$ .
The original number is $6$ , but its proposed partner is
$3-\sqrt9=0$

Multiplying by that partner over itself would mean $0/0$ , so it is invalid.
Requiring $d$ to be a positive integer that is not a perfect square makes $\sqrt d$ irrational and restores the nonzero-partner guarantee for a nonzero original number.

Answer

The claim fails: $a+b\sqrt d=6$ , $a-b\sqrt d=0$ ; require $d$ to be a positive nonsquare integer.

Key idea

The nonsquare condition protects the nonzero multiplier used in radical conjugation.
Problem 9: Mark complete
Problem 10 Closed under division

Let $d$ be a positive integer that is not a perfect square, and let $a$ , $b$ , $c$ and $e$ be rational, with $c+e\sqrt d\ne0$ . Find rational $p$ and $q$ with $\dfrac{a+b\sqrt d}{c+e\sqrt d}=p+q\sqrt d$ , and explain why both coefficients are defined and rational.
1. Hint 1
  Multiply numerator and denominator by the partner $c-e\sqrt d$ .
2. Hint 2
  Before dividing, show that the new denominator $c^2-e^2d$ is not zero; treat $e\ne0$ and $e=0$ separately.
Answer

$p=\frac{ac-bed}{c^2-e^2d}$ and $q=\frac{bc-ae}{c^2-e^2d}$ , where $c^2-e^2d\ne0$ .

Full solution

Multiplying by $c-e\sqrt d$ over itself, the denominator becomes
$c^2-e^2d$
and the numerator becomes
$(ac-bed)+(bc-ae)\sqrt d$

The new denominator is not zero.
If $e\ne0$ and $c^2=e^2d$ , then $\sqrt d=|c/e|$ would be rational, but a positive integer that is not a perfect square has an irrational square root.
If $e=0$ , the condition $c+e\sqrt d\ne0$ gives $c\ne0$ , so $c^2\ne0$ .

So the partner is not zero either, multiplying by it over itself keeps the value, and $p$ and $q$ are rational numbers divided by a nonzero rational number, which makes both rational.

Answer

$p=\frac{ac-bed}{c^2-e^2d}$ and $q=\frac{bc-ae}{c^2-e^2d}$ , where $c^2-e^2d\ne0$ .

Key idea

Numbers of the form $p+q\sqrt d$ are closed under division by a nonzero one, because the partner makes the denominator rational.
Problem 10: Mark complete

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