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A bad title I know. I'm having trouble understanding the answer to the following question:

Solve the equation $\sqrt{6+3\sqrt2} = \sqrt a + \sqrt b$, writing a and b in the form $a + b\sqrt c$.

The answer starts by squaring both sides and then equating the parts with square roots.

$a + b = 6$ and $2\sqrt{ab} = 3\sqrt2$

I don't understand why they assume that $a + b$ is equal to $6$ because neither $a$ nor $b$ appear to be rational.

Are you supposed to assume that $a$ and $b$ have the same $b\sqrt c$ term because it doesn't ask for multiple solutions?

The answers they give are: $a=3\pm\dfrac{3\sqrt2}{2}$, $b=3\pm\dfrac{3\sqrt2}{2}$

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    $\begingroup$ The question seems to be badly posed, obviously one possible answer is $a=6+3\sqrt2$, $b=0$. $\endgroup$ – David Nov 22 '17 at 1:21
  • $\begingroup$ This is just a question in a practise paper so they're not too concerned with correctness. I've added the answers they gave. $\endgroup$ – Zac Pullar-Strecker Nov 22 '17 at 1:26
  • $\begingroup$ @ZacPullar-Strecker As David points out the question is badly posed. There are many correct answers, not just the two they list. I think you should ask your teacher for clarification. $\endgroup$ – Malcolm Nov 22 '17 at 1:43
  • $\begingroup$ @Malcolm Yeah I see what you mean, thanks! $\endgroup$ – Zac Pullar-Strecker Nov 22 '17 at 1:52
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If $a$ and $b$ are required to be integers, then you can equate the parts.

If not, then, for any $6+3\sqrt2 \ge a> 0$, you can set $b =(\sqrt{6+3\sqrt2} - \sqrt a)^2 $.

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