Rationalizing and Radical Conjugates: Practice

The radical conjugate flips the sign of the surd part and leaves the rational part alone. Here the rational part is $3$ and the surd part is $+\sqrt{5}$.

$(3 + \sqrt{5})^{*} = 3 - \sqrt{5}$

Nothing else changes. Negating the whole expression would give $-3 - \sqrt{5}$, and negating only the rational part would give $\sqrt{5} - 3$; neither is the conjugate.

You want the factor that turns $\sqrt{7}$ into a whole power of $7$. Multiplying a square root by itself does exactly that.

$\sqrt{7}\cdot\sqrt{7} = 7$

So multiply top and bottom by $\sqrt{7}$, giving $\frac{2\sqrt{7}}{7}$. The other three all leave a radical downstairs: $7\sqrt{7}$, $2\sqrt{7}$, and $\sqrt{14}$ are all irrational.

Multiply top and bottom by the radical conjugate $2 - \sqrt{3}$, and the denominator collapses to a difference of squares.

$(2 + \sqrt{3})(2 - \sqrt{3}) = 4 - 3 = 1$

So the fraction becomes $\frac{2 - \sqrt{3}}{1} = 2 - \sqrt{3}$. The denominator of $1$ is not a mistake: $2 + \sqrt{3} \approx 3.732$ and $2 - \sqrt{3} \approx 0.268$, and those two numbers really do multiply to $1$.

Multiply top and bottom by $\sqrt{2}$, which is multiplying by $1$.

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

The numerator picks up the radical and the denominator becomes $2$. Numerically, $3 \div 1.414 \approx 2.121$, and $3(1.414) \div 2 \approx 2.121$, so the value is unchanged.

Multiply top and bottom by $\sqrt{3}$ to clear the denominator.

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

The last step is the one people forget: $6$ and $3$ share a factor of $3$, so the fraction is not finished until you cancel it. As a check, $6 \div 1.732 \approx 3.464$, and $2(1.732) \approx 3.464$.

The denominator is a two-term surd, so multiply top and bottom by its conjugate $\sqrt{3} + 1$.

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

That leaves $\frac{\sqrt{3} + 1}{2}$. Checking the size, $\sqrt{3} - 1 \approx 0.732$ and $1 \div 0.732 \approx 1.366$, which matches $(1.732 + 1) \div 2 \approx 1.366$.

Expand the product and watch the cross terms cancel.

$(a + b\sqrt{d})(a - b\sqrt{d}) = a^2 - ab\sqrt{d} + ab\sqrt{d} - b^2(\sqrt{d})^2 = a^2 - b^2 d$

Two details carry all the weight. The coefficient $b$ is squared, not left alone, and $(\sqrt{d})^2$ becomes the rational number $d$, so no radical survives.

Multiply top and bottom by $\sqrt{6}$.

$\frac{4}{\sqrt{6}}\cdot\frac{\sqrt{6}}{\sqrt{6}} = \frac{4\sqrt{6}}{6} = \frac{2\sqrt{6}}{3}$

Now reduce: $4$ and $6$ share a factor of $2$. As a check, $4 \div 2.449 \approx 1.633$, and $2(2.449) \div 3 \approx 1.633$.

The cross terms cancel, leaving the difference of the squares in the order they appear.

$(\sqrt{7} + 3)(\sqrt{7} - 3) = (\sqrt{7})^2 - 3^2 = 7 - 9 = -2$

The answer is negative, and that is perfectly fine. The conjugate product is guaranteed to be rational, not to be positive.

Simplify the radical first, since $\sqrt{8} = 2\sqrt{2}$, and then clear it.

$\frac{1}{\sqrt{8}} = \frac{1}{2\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{4}$

Simplifying first keeps the numbers small. Multiplying by $\sqrt{8}$ instead gives $\frac{\sqrt{8}}{8}$, which reduces to the same $\frac{\sqrt{2}}{4}$ after you replace $\sqrt{8}$ with $2\sqrt{2}$. Numerically both are about $0.354$.

Rationalize the second option and see what comes out. Its conjugate is $\sqrt{3} + \sqrt{2}$, and the conjugate product is $3 - 2 = 1$.

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

So $\sqrt{3} - \sqrt{2}$ and $\sqrt{3} + \sqrt{2}$ are reciprocals of each other. Note that $\sqrt{3} + \sqrt{2} \approx 3.146$ while $\sqrt{5} \approx 2.236$: square roots do not add like that.

The cross terms cancel and the squares remain.

$(1 + \sqrt{2})(1 - \sqrt{2}) = 1^2 - (\sqrt{2})^2 = 1 - 2 = -1$

Since the product is $-1$, the two numbers are negative reciprocals: $\frac{1}{1 + \sqrt{2}} = -(1 - \sqrt{2}) = \sqrt{2} - 1$.