0606 P23 - Jun 2023 - Q2 - 4 marks
7705
Do not use a calculator in this question.
Write
\(\frac{\sqrt{98x^{12}}}{3+\sqrt2}\)
in the form
\((a\sqrt b+c)x^d,\)
where \(a\), \(b\), \(c\) and \(d\) are integers.
Solution
Answer: \((3\sqrt2-2)x^6\).
Rewrite the equation as a standard quadratic in the chosen variable, then solve and reject any value that does not satisfy the original form.
First simplify the surd in the numerator:
\(\sqrt{98x^{12}}=\sqrt{49\cdot2\cdot x^{12}}=7\sqrt2\,x^6.\)
So
\(\frac{\sqrt{98x^{12}}}{3+\sqrt2} =\frac{7\sqrt2\,x^6}{3+\sqrt2}.\)
Rationalise the denominator:
\(\frac{7\sqrt2\,x^6}{3+\sqrt2}\cdot\frac{3-\sqrt2}{3-\sqrt2} =\frac{7\sqrt2(3-\sqrt2)}{9-2}x^6.\)
Since the denominator is \(7\), this becomes
\(\sqrt2(3-\sqrt2)x^6.\)
Therefore
\(\frac{\sqrt{98x^{12}}}{3+\sqrt2} =(3\sqrt2-2)x^6.\)
