Rationalizing and Radical Conjugates: Additional Practice (Set 2)

Use the sum-of-cubes multiplier $x^2 - xy + y^2$ with $x = \sqrt[3]{4}$ and $y = 1$. Here $x^2 = \sqrt[3]{16} = 2\sqrt[3]{2}$, so the multiplier is $2\sqrt[3]{2} - \sqrt[3]{4} + 1$.

$\left(\sqrt[3]{4} + 1\right)\left(2\sqrt[3]{2} - \sqrt[3]{4} + 1\right) = \left(\sqrt[3]{4}\right)^3 + 1^3 = 4 + 1 = 5$

The answer is $\frac{2\sqrt[3]{2} - \sqrt[3]{4} + 1}{5}$, about $1.932 \div 5 \approx 0.386$, which matches $1 \div 2.587 \approx 0.386$.

Use the sum-of-cubes multiplier $x^2 - xy + y^2$ with $x = \sqrt[3]{7}$ and $y = 1$, which is $\sqrt[3]{49} - \sqrt[3]{7} + 1$.

$\left(\sqrt[3]{7} + 1\right)\left(\sqrt[3]{49} - \sqrt[3]{7} + 1\right) = \left(\sqrt[3]{7}\right)^3 + 1^3 = 7 + 1 = 8$

So the answer is $\frac{\sqrt[3]{49} - \sqrt[3]{7} + 1}{8} \approx 2.746 \div 8 \approx 0.343$, which matches $1 \div 2.913 \approx 0.343$.

Rationalize the numerator by multiplying top and bottom by $\sqrt{n+1} + \sqrt{n}$.

$\sqrt{n+1} - \sqrt{n} = \frac{(n+1) - n}{\sqrt{n+1} + \sqrt{n}} = \frac{1}{\sqrt{n+1} + \sqrt{n}}$

That identity explains why consecutive square roots crowd together: the denominator grows without bound while the numerator stays at $1$. It is exact, unlike $\frac{1}{2\sqrt{n}}$, which at $n = 1$ gives $0.5$ while the true difference is about $0.414$.

The shorter leg is $1$ and the hypotenuse is $\sqrt{1^2 + (\sqrt{2})^2} = \sqrt{3}$, so the ratio starts as $\frac{1}{\sqrt{3}}$. Clear the radical by multiplying top and bottom by $\sqrt{3}$.

$\frac{1}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}$

The value is about $0.577$, which is less than $1$ as any leg-to-hypotenuse ratio must be.

Apply the identity $(a + b\sqrt{d})(a - b\sqrt{d}) = a^2 - b^2 d$ with $a = 7$ and $b = 2$.

$49 - 4d = 25 \;\Longrightarrow\; 4d = 24 \;\Longrightarrow\; d = 6$

Check it: $(7 + 2\sqrt{6})(7 - 2\sqrt{6}) = 49 - 4(6) = 25$. Note that $b = 2$ contributes $b^2 = 4$, not $2$; using $49 - 2d = 25$ would give the wrong answer $d = 12$.

Rationalize the fraction before combining anything.

$\frac{6}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3}$

Now the two terms are like terms, so subtract their coefficients:

$2\sqrt{3} - \sqrt{3} = \sqrt{3} \approx 1.732$

The third option is a trap: $\frac{6 - \sqrt{3}}{\sqrt{3}} = 2\sqrt{3} - 1 \approx 2.464$, because subtracting $\sqrt{3}$ from the numerator is not the same as subtracting it from the whole fraction.

The conjugate product is $a^2 - (\sqrt{7})^2 = a^2 - 7$, so set that equal to $2$.

$a^2 - 7 = 2 \;\Longrightarrow\; a^2 = 9 \;\Longrightarrow\; a = 3$

Taking the positive root gives $a = 3$, and indeed $(3 + \sqrt{7})(3 - \sqrt{7}) = 9 - 7 = 2$. The value $a = 9$ would give $81 - 7 = 74$, and $a = 2$ would give $4 - 7 = -3$.

Multiply top and bottom by the conjugate $7 + 4\sqrt{3}$, remembering to square the coefficient $4$.

$(7 - 4\sqrt{3})(7 + 4\sqrt{3}) = 49 - 16(3) = 49 - 48 = 1$

So the reciprocal is $7 + 4\sqrt{3}$, giving $a = 7$ and $b = 4$, and therefore $ab = 28$. Check the size: $7 - 4\sqrt{3} \approx 0.0718$ and $1 \div 0.0718 \approx 13.93 \approx 7 + 4(1.732)$.

Multiply top and bottom by the conjugate $\sqrt{x} + 1$. The denominator becomes $(\sqrt{x})^2 - 1^2 = x - 1$, which is nonzero because $x > 1$.

$\frac{\sqrt{x}}{\sqrt{x} - 1}\cdot\frac{\sqrt{x} + 1}{\sqrt{x} + 1} = \frac{x + \sqrt{x}}{x - 1}$

The numerator distributes as $\sqrt{x}\cdot\sqrt{x} + \sqrt{x}\cdot 1 = x + \sqrt{x}$. Test it at $x = 4$: the original is $\frac{2}{1} = 2$, and the answer gives $\frac{4 + 2}{3} = 2$.

Rationalize each numerator with the identity $\sqrt{n+1} - \sqrt{n} = \frac{1}{\sqrt{n+1} + \sqrt{n}}$, so that comparing the differences becomes comparing denominators.

$\sqrt{2} - \sqrt{1} = \frac{1}{\sqrt{2} + 1} \approx \frac{1}{2.414}, \qquad \sqrt{1001} - \sqrt{1000} \approx \frac{1}{63.2}$

The smallest denominator gives the largest fraction, and the denominators grow with $n$. So the gap shrinks as $n$ grows, and the largest difference is the one at $n = 1$, about $0.414$. No calculator is needed: only the direction of the trend.

Multiply top and bottom by the conjugate $\sqrt{x} - \sqrt{y}$. The denominator becomes a difference of squares, and $x \ne y$ guarantees it is not zero.

$(\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y}) = x - y$

So the answer is $\frac{\sqrt{x} - \sqrt{y}}{x - y}$. Test it at $x = 4$ and $y = 1$: the original is $\frac{1}{2 + 1} = \frac{1}{3}$, and the answer gives $\frac{2 - 1}{3} = \frac{1}{3}$. The condition $x \ne y$ matters: if $x = y$ the conjugate is $0$, and multiplying by $\frac{0}{0}$ is not multiplying by $1$.

Combine over a common denominator, and notice that $x$ and $y$ are conjugates, so their product is rational.

$\frac{1}{x} - \frac{1}{y} = \frac{y - x}{xy}, \qquad xy = 7 - 5 = 2, \qquad y - x = 2\sqrt{5}$

Putting those together, the $2$s cancel:

$\frac{2\sqrt{5}}{2} = \sqrt{5} \approx 2.236$

Rationalizing each fraction separately gives the same thing: $\frac{1}{x} = \frac{\sqrt{7} + \sqrt{5}}{2}$ and $\frac{1}{y} = \frac{\sqrt{7} - \sqrt{5}}{2}$, whose difference is $\sqrt{5}$.