Rationalizing and Radical Conjugates: Additional Practice (Set 1)

Find the reciprocal by multiplying by the conjugate. Since $(3 + 2\sqrt{2})(3 - 2\sqrt{2}) = 9 - 4(2) = 1$, the reciprocal is the conjugate itself.

$\frac{1}{x} = 3 - 2\sqrt{2}$

Adding a number to its conjugate cancels the surd parts and leaves twice the rational part:

$(3 + 2\sqrt{2}) + (3 - 2\sqrt{2}) = 6$

Rationalize each fraction separately. Both conjugate products equal $3 - 1 = 2$, and the $2$ upstairs cancels it each time.

$\frac{2}{\sqrt{3} + 1} = \frac{2(\sqrt{3} - 1)}{2} = \sqrt{3} - 1, \qquad \frac{2}{\sqrt{3} - 1} = \frac{2(\sqrt{3} + 1)}{2} = \sqrt{3} + 1$

Adding, the $1$s cancel and the surds double:

$(\sqrt{3} - 1) + (\sqrt{3} + 1) = 2\sqrt{3} \approx 3.464$

Multiply top and bottom by the conjugate $\sqrt{6} + \sqrt{5}$. The conjugate product subtracts, not adds.

$(\sqrt{6} - \sqrt{5})(\sqrt{6} + \sqrt{5}) = 6 - 5 = 1$

Since the denominator is $1$, the answer is $\sqrt{6} + \sqrt{5} \approx 4.686$. Sanity check: $\sqrt{6} - \sqrt{5} \approx 0.213$, and $1 \div 0.213 \approx 4.686$. Two consecutive roots are close together, so their reciprocal difference is large.

The denominator is $5^{1/3}$, so the factor that completes the cube is $5^{2/3} = \sqrt[3]{25}$.

$\frac{\sqrt[3]{2}}{\sqrt[3]{5}}\cdot\frac{\sqrt[3]{25}}{\sqrt[3]{25}} = \frac{\sqrt[3]{2 \cdot 25}}{5} = \frac{\sqrt[3]{50}}{5}$

Multiplying by $\sqrt[3]{5}$ instead would leave $\sqrt[3]{25}$ downstairs. Numerically the original is $1.260 \div 1.710 \approx 0.737$, and $3.684 \div 5 \approx 0.737$.

Rationalize the original. Since $(\sqrt{5} - 2)(\sqrt{5} + 2) = 5 - 4 = 1$, the reciprocal is the conjugate.

$\frac{1}{\sqrt{5} - 2} = \sqrt{5} + 2 \approx 4.236$

The third and fourth options are the same number in disguise, because $2\sqrt{5} - 4 = 2(\sqrt{5} - 2)$ and $4\sqrt{5} - 8 = 4(\sqrt{5} - 2)$, so each fraction reduces back to $\frac{1}{\sqrt{5} - 2}$. Only $\sqrt{5} - 2 \approx 0.236$ is different: it is the reciprocal of the answer, not the answer.

Multiply top and bottom by the conjugate $\sqrt{3} + \sqrt{2}$. The denominator becomes $3 - 2 = 1$, and the numerator becomes a square.

$(\sqrt{3} + \sqrt{2})^2 = 3 + 2\sqrt{6} + 2 = 5 + 2\sqrt{6}$

With a denominator of $1$, that is the whole answer, about $9.899$. Compare it with the direct division: $3.146 \div 0.318 \approx 9.899$. Dropping the middle term $2\sqrt{6}$ of the square is the usual mistake.

Rationalize each term. Using $\sqrt{8} = 2\sqrt{2}$ and $\sqrt{32} = 4\sqrt{2}$, every term is a multiple of $\sqrt{2}$.

$\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}, \qquad \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4}, \qquad \frac{1}{4\sqrt{2}} = \frac{\sqrt{2}}{8}$

Factor out $\sqrt{2}$ and add the fractions:

$\sqrt{2}\left(\frac{1}{2} + \frac{1}{4} + \frac{1}{8}\right) = \sqrt{2}\cdot\frac{7}{8} = \frac{7\sqrt{2}}{8} \approx 1.237$

Multiply top and bottom by the conjugate $\sqrt{5} - \sqrt{2}$. The denominator becomes $5 - 2 = 3$, which cancels the $3$ on top.

$\frac{3(\sqrt{5} - \sqrt{2})}{3} = \sqrt{5} - \sqrt{2}$

Matching this against $a\sqrt{5} + b\sqrt{2}$ gives $a = 1$ and $b = -1$, so $a = 1$. The value is about $0.822$, matching $3 \div 3.650$.

Regroup before expanding anything. The two squares can be merged into the square of one product.

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

The inside is a conjugate product, worth $4 - 3 = 1$, so the whole expression is $1^2 = 1$.

$\left[1\right]^2 = 1$

Expanding each square first gives $(7 + 4\sqrt{3})(7 - 4\sqrt{3}) = 49 - 48 = 1$, the same answer through more work.

Multiply top and bottom by the conjugate and read off the shape of the result.

$\frac{1}{a + b\sqrt{d}} = \frac{a - b\sqrt{d}}{a^2 - b^2 d} = \frac{a}{a^2 - b^2 d} + \left(\frac{-b}{a^2 - b^2 d}\right)\sqrt{d}$

Both coefficients are quotients of rationals, so they are rational, and the family is closed under division. The other options fail: $a^2 - b^2 d$ vanishes only when $a = b = 0$, which is excluded, and it is not always positive, since $a = b = 1$ and $d = 2$ give $1 - 2 = -1$.

Multiply top and bottom by the conjugate $\sqrt{5} - \sqrt{3}$. The denominator becomes $5 - 3 = 2$, and the numerator becomes a square.

$(\sqrt{5} - \sqrt{3})^2 = 5 - 2\sqrt{15} + 3 = 8 - 2\sqrt{15}$

Every term now shares a factor of $2$, so cancel it against the denominator:

$\frac{8 - 2\sqrt{15}}{2} = 4 - \sqrt{15} \approx 0.127$

That matches the direct division $0.504 \div 3.968 \approx 0.127$. Two of the wrong options are the two ways the cancellation goes astray. Using $5 + 3 = 8$ for the conjugate product, instead of $5 - 3$, leaves $\frac{8 - 2\sqrt{15}}{8}$, and cancelling the factor of $2$ a second time leaves $\frac{4 - \sqrt{15}}{2}$.

Expand the product and watch every intermediate term cancel in pairs.

$(x + y)(x^2 - xy + y^2) = x^3 - x^2y + xy^2 + x^2y - xy^2 + y^3 = x^3 + y^3$

Only the cubes are left, and a cube is exactly what turns a cube root rational. The multiplier is not a conjugate and not a square; it is engineered so that the surviving powers are the ones that land back in the rational numbers.