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# Rational Roots

David sent in this solution, using the hints we gave you.

$\sqrt{a}+\sqrt{b}$ rational

$\Rightarrow (\sqrt{a}+\sqrt{b})^2=a+b+2\sqrt{a b}$ rational

$\Rightarrow 2\sqrt{a b}$ rational

$\Rightarrow \sqrt{a b}$ rational

$\Rightarrow a+\sqrt{a b}$ rational

i.e., $\sqrt{a}(\sqrt{a}+\sqrt{b})$ rational

$\Rightarrow \sqrt{a}$ rational (and so $a$ is a square)

$\Rightarrow \sqrt{b}$ is also rational and hence $b$ is a square.

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David sent in this solution, using the hints we gave you.

$\sqrt{a}+\sqrt{b}$ rational

$\Rightarrow (\sqrt{a}+\sqrt{b})^2=a+b+2\sqrt{a b}$ rational

$\Rightarrow 2\sqrt{a b}$ rational

$\Rightarrow \sqrt{a b}$ rational

$\Rightarrow a+\sqrt{a b}$ rational

i.e., $\sqrt{a}(\sqrt{a}+\sqrt{b})$ rational

$\Rightarrow \sqrt{a}$ rational (and so $a$ is a square)

$\Rightarrow \sqrt{b}$ is also rational and hence $b$ is a square.

For the second part:

$\sqrt{a}+\sqrt{b}+\sqrt{c}$ rational

$\Rightarrow (\sqrt{a}+\sqrt{b}+\sqrt{c})^2$ rational

$\Rightarrow \sqrt{a b}+\sqrt{b c}+\sqrt{c a}$ rational

$(\sqrt{a b}+\sqrt{b c}+\sqrt{c a})^2=a b+b c+c a+2\sqrt{a b
c}(\sqrt{a}+\sqrt{b}+ \sqrt{c})$

so $\sqrt{a b c}$ is also rational

$\Rightarrow \sqrt{a}(\sqrt{a b}+\sqrt{b c}+\sqrt{c
a})-\sqrt{a b c}=a(\sqrt{b} +\sqrt{c})$ is rational

so $\sqrt{b}+\sqrt{c}$ is rational and $\sqrt{a}$ is rational,
and use the above.

Solve quadratic equations and use continued fractions to find rational approximations to irrational numbers.