Review question

# If $yz=a^2$, can we prove that $1/(a+y) + 1/(a+z) = 1/a$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6806

## Solution

If $yz=a^2$ prove that $\frac{1}{a+y} + \frac{1}{a+z} = \frac{1}{a}.$

Solving $yz=a^2$ for $y$, we obtain

$y=\dfrac{a^2}{z},$

which we can substitute into the left-hand side of the given expression. This leads us to

\begin{align*} \frac{1}{a+y} + \frac{1}{a+z} &= \frac{1}{a+\dfrac{a^2}{z}} + \frac{1}{a+z}\\ &=\dfrac{z}{az+a^2} + \dfrac{a}{a(a+z)}\\ &=\dfrac{a+z}{a(a+z)}=\dfrac{1}{a}, \end{align*}

as required.

By using this result, or otherwise, find the numerical values of

1. $\dfrac{1}{1+x^k} + \dfrac{1}{1+x^{-k}}$;

Comparing this expression with the one above leads to setting $a=1$, $x^k=y$ and $x^{-k}=z$. These values clearly fulfil $yz=a^2$, and hence give us

$\frac{1}{1+x^k} + \frac{1}{1+x^{-k}} = 1.$

1. $\dfrac{1}{7+\sqrt{62}-\sqrt{13}} + \dfrac{1}{7+\sqrt{62}+\sqrt{13}}$;

We have $(\sqrt{62}-\sqrt{13})(\sqrt{62}+\sqrt{13})=62-13=49=7^2$, which means we can we take $a=7$, $y=\sqrt{62}-\sqrt{13}$ and $z=\sqrt{62}+\sqrt{13}$.

Thus $\frac{1}{7+\sqrt{62}-\sqrt{13}} + \frac{1}{7+\sqrt{62}+\sqrt{13}}=\frac{1}{7}.$

1. $\dfrac{2}{5(2+5\log_b c)} + \dfrac{1}{5+2\log_c b}$;

We can write this as

$\frac{1}{5+\dfrac{25}{2}\log_b c} + \frac{1}{5+2\log_c b}.$

This looks rather like the expression at the start of the question, so let us try $a=5$, $y=\dfrac{25}{2}\log_b c$ and $z=2\log_c b$. Then $yz=\dfrac{25}{2}\log_b c\times2\log_c b = 25\log_b c\times\log_c b = 25 = a^2 .$

We have used the law of logarithms for changing bases which states that $\log_a b=\log_a c\times \log_c b .$

So we can state that $\frac{2}{5(2+5\log_b c)} + \frac{1}{5+2\log_c b}=\frac{1}{5}.$