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Surds and Rationalization of Denominators

25 questions

All types25Multiple choice16Numeric9
Question 21Question →

If 6+26−2=p+q3\frac{\sqrt{6} + \sqrt{2}}{\sqrt{6} - \sqrt{2}} = p + q\sqrt{3}, where pp and qq are rational numbers, what is the value of p+qp + q?

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Answer: 3

Answer

The value of p+qp + q is 3.
Multiplying both numerator and denominator by the conjugate (6+2)(\sqrt{6} + \sqrt{2}) yields 8+434=2+3\frac{8 + 4\sqrt{3}}{4} = 2 + \sqrt{3}. Matching terms with p+q3p + q\sqrt{3} gives p=2p = 2 and q=1q = 1, so p+q=3p + q = 3.

Step-by-Step Solution

1
Multiply the numerator and denominator by the conjugate of the denominator, (6+2)(\sqrt{6} + \sqrt{2}).
(6+2)(6+2)(6−2)(6+2)\frac{(\sqrt{6} + \sqrt{2})(\sqrt{6} + \sqrt{2})}{(\sqrt{6} - \sqrt{2})(\sqrt{6} + \sqrt{2})}
To eliminate radicals from the denominator.
2
Expand both numerator and denominator.
Numerator: 6+212+2=8+436 + 2\sqrt{12} + 2 = 8 + 4\sqrt{3}. Denominator: 6−2=46 - 2 = 4.
Using algebraic expansion (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2 and difference of two squares.
3
Simplify the resulting fraction.
8+434=2+3\frac{8 + 4\sqrt{3}}{4} = 2 + \sqrt{3}
Dividing each term in the numerator by 4.
4
Equate to p+q3p + q\sqrt{3} to determine pp and qq, then find p+qp + q.
p=2p = 2, q=1  ⟹  p+q=2+1=3q = 1 \implies p + q = 2 + 1 = 3.
Comparing rational and irrational parts separately.

Key Concept

Rationalization of Denominators and Surd Conjugates
Question 22Question →

What is the simplified numerical value of 123+1+123−1\frac{\sqrt{12}}{\sqrt{3} + 1} + \frac{\sqrt{12}}{\sqrt{3} - 1}?

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Answer: 6

Answer

The simplified numerical value of the given expression is 6.
Combining the fractions using their common conjugate denominator (3+1)(3−1)=2(\sqrt{3}+1)(\sqrt{3}-1) = 2 leads to a numerator of 23[(3−1)+(3+1)]=23(23)=122\sqrt{3}[(\sqrt{3}-1)+(\sqrt{3}+1)] = 2\sqrt{3}(2\sqrt{3}) = 12. Dividing 12 by 2 yields the final answer of 6.

Step-by-Step Solution

1
Simplify the surd in the numerator
\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}
Simplifying surds into basic form makes subsequent calculations simpler.
2
Combine the fractions by finding a common denominator
23(3−1)+23(3+1)(3+1)(3−1)\frac{2\sqrt{3}(\sqrt{3}-1) + 2\sqrt{3}(\sqrt{3}+1)}{(\sqrt{3}+1)(\sqrt{3}-1)}
Multiplying denominators forms a conjugate pair, which rationalizes the combined denominator.
3
Simplify the numerator and the denominator
\text{Numerator} = 2\sqrt{3}(\sqrt{3}-1 + \sqrt{3}+1) = 2\sqrt{3}(2\sqrt{3}) = 12; \quad \text{Denominator} = (\sqrt{3})^2 - 1^2 = 3 - 1 = 2
Expanding the numerator combines like surd terms, and using the difference of squares simplifies the denominator to a rational integer.
4
Perform the final division
122=6\frac{12}{2} = 6
Dividing the simplified numerator by the rationalized denominator gives the final integer value.

Key Concept

Rationalization of binomial denominators using conjugate surds
Question 23Question →

A rectangle has an area of 10 cm210\text{ cm}^2 and a length of (7+2) cm(\sqrt{7} + \sqrt{2})\text{ cm}. What is the width of the rectangle in simplified surd form?

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Answer: (27−22) cm(2\sqrt{7} - 2\sqrt{2})\text{ cm}

Answer

The width of the rectangle in simplified surd form is (27−22) cm(2\sqrt{7} - 2\sqrt{2})\text{ cm}.
The area formula for a rectangle gives width=107+2\text{width} = \frac{10}{\sqrt{7} + \sqrt{2}}. Multiplying both numerator and denominator by the conjugate (7−2)(\sqrt{7} - \sqrt{2}) produces 10(7−2)7−2=10(7−2)5=27−22\frac{10(\sqrt{7} - \sqrt{2})}{7 - 2} = \frac{10(\sqrt{7} - \sqrt{2})}{5} = 2\sqrt{7} - 2\sqrt{2}.

Step-by-Step Solution

1
Set up the formula for the width of the rectangle
\text{Width} = \frac{\text{Area}}{\text{Length}} = \frac{10}{\sqrt{7} + \sqrt{2}}
The area of a rectangle is length multiplied by width.
2
Rationalize the denominator by multiplying top and bottom by the conjugate (7−2)(\sqrt{7} - \sqrt{2})
\text{Width} = \frac{10(\sqrt{7} - \sqrt{2})}{(\sqrt{7} + \sqrt{2})(\sqrt{7} - \sqrt{2})}
Multiplying by the conjugate creates a difference of squares in the denominator, eliminating radicals.
3
Simplify the denominator using (a+b)(a−b)=a2−b2(a+b)(a-b) = a^2 - b^2
(\sqrt{7})^2 - (\sqrt{2})^2 = 7 - 2 = 5
Squaring a square root yields the underlying rational number.
4
Divide the numerator by the simplified denominator
\frac{10(\sqrt{7} - \sqrt{2})}{5} = 2(\sqrt{7} - \sqrt{2}) = 2\sqrt{7} - 2\sqrt{2}
Dividing 1010 by 55 gives 22, which is then distributed across the terms inside the parentheses.

Key Concept

Rationalization of Denominators with Binomial Surds
Question 24Question →

If 5+35−3−5−35+3=k15\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} - \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}} = k\sqrt{15}, find the value of kk.

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Answer: 2

Answer

The value of kk is 22.
Combining the fractions gives a common denominator of (5−3)(5+3)=5−3=2(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3}) = 5-3 = 2. Expanding the numerator gives (8+215)−(8−215)=415(8+2\sqrt{15}) - (8-2\sqrt{15}) = 4\sqrt{15}. Dividing by 2 yields 2152\sqrt{15}, giving k=2k = 2.

Step-by-Step Solution

1
Combine the fractions on the left-hand side over a common denominator.
(5+3)2−(5−3)2(5−3)(5+3)\frac{(\sqrt{5} + \sqrt{3})^2 - (\sqrt{5} - \sqrt{3})^2}{(\sqrt{5} - \sqrt{3})(\sqrt{5} + \sqrt{3})}
Combining two fractions with conjugate denominators simplifies the expression.
2
Evaluate the denominator using the difference of two squares formula (a−b)(a+b)=a2−b2(a - b)(a + b) = a^2 - b^2.
(\sqrt{5})^2 - (\sqrt{3})^2 = 5 - 3 = 2
Multiplying conjugate surds eliminates the radical signs in the denominator.
3
Expand both squared terms in the numerator and subtract them.
(8 + 2\sqrt{15}) - (8 - 2\sqrt{15}) = 4\sqrt{15}
Expanding (a±b)2=a2±2ab+b2(a \pm b)^2 = a^2 \pm 2ab + b^2 gives 5±215+3=8±2155 \pm 2\sqrt{15} + 3 = 8 \pm 2\sqrt{15}.
4
Divide the resulting numerator by the denominator and solve for kk.
\frac{4\sqrt{15}}{2} = 2\sqrt{15} \Rightarrow k = 2
Dividing 4154\sqrt{15} by 22 yields 2152\sqrt{15}, so matching the coefficients gives k=2k = 2.

Key Concept

Rationalization of Denominators and Difference of Conjugate Surd Fractions
Estimated Time:1m 30s
Question 25Question →

What is the simplified form of the expression 32+233−2\frac{3\sqrt{2} + 2\sqrt{3}}{\sqrt{3} - \sqrt{2}}?

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Answer: 12+5612 + 5\sqrt{6}

Answer

The simplified form is 12+5612 + 5\sqrt{6}.
Multiplying both numerator and denominator by the conjugate of the denominator, (3+2)(\sqrt{3} + \sqrt{2}), clears the square roots in the denominator (resulting in 3−2=13 - 2 = 1). Expanding the numerator gives 36+6+6+263\sqrt{6} + 6 + 6 + 2\sqrt{6}, which simplifies cleanly to 12+5612 + 5\sqrt{6}.

Step-by-Step Solution

1
Identify the conjugate of the denominator
The conjugate of (3−2)(\sqrt{3} - \sqrt{2}) is (3+2)(\sqrt{3} + \sqrt{2}).
To rationalize a binomial denominator of the form (a−b)(\sqrt{a} - \sqrt{b}), multiply numerator and denominator by (a+b)(\sqrt{a} + \sqrt{b}).
2
Multiply the numerator and denominator by the conjugate
(32+23)(3+2)(3−2)(3+2)\frac{(3\sqrt{2} + 2\sqrt{3})(\sqrt{3} + \sqrt{2})}{(\sqrt{3} - \sqrt{2})(\sqrt{3} + \sqrt{2})}
This removes the radical terms from the denominator using the difference of squares identity.
3
Expand the numerator and simplify the denominator
Denominator: (3)2−(2)2=3−2=1(\sqrt{3})^2 - (\sqrt{2})^2 = 3 - 2 = 1.
Numerator: 32⋅3+32⋅2+23⋅3+23⋅2=36+6+6+263\sqrt{2}\cdot\sqrt{3} + 3\sqrt{2}\cdot\sqrt{2} + 2\sqrt{3}\cdot\sqrt{3} + 2\sqrt{3}\cdot\sqrt{2} = 3\sqrt{6} + 6 + 6 + 2\sqrt{6}.
Apply the distributive property and basic radical simplification rules.
4
Combine like terms in the numerator
(6+6)+(36+26)=12+56(6 + 6) + (3\sqrt{6} + 2\sqrt{6}) = 12 + 5\sqrt{6}.
Collect rational numbers together and like surd terms together.

Key Concept

Rationalization of Denominators with Binomial Surds
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