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Mathematics · 2026 · 2 marks
CBSE 2026 · Region 4 · Set 1 · Q24
(a) Prove that 2+3 √ 5 is an irrational number given that √ 5 is irrational number.
(a)
Prove that $\displaystyle 2+3 \sqrt{5}$ is an irrational number given that $\displaystyle \sqrt{5}$ is irrational number.
OR
(b)
If the HCF of $\displaystyle 210$ and $\displaystyle 55$ is expressed as $\displaystyle 210$ × $\displaystyle 5$ + 55m, then find the value of m.
Marking-scheme solution
Let \(\displaystyle 2+3 \sqrt{5}\) be a rational number.
\(\displaystyle \therefore 2+3 \sqrt{5}=\frac{\mathrm{p}}{\mathrm{q}}\), where \(\displaystyle \mathrm{q} \neq 0\) and p and q are integers.
\(\displaystyle \Rightarrow \sqrt{5}=\frac{p-2 q}{3 q}\)
As \(\displaystyle \frac{p-2 q}{3 q}\) is a rational number, so \(\displaystyle \sqrt{5}\) is rational.
But we know that \(\displaystyle \sqrt{5}\) is irrational.
∴ Our assumption is wrong. Hence, \(\displaystyle 2+3 \sqrt{5}\) is an irrational number.
\(\displaystyle 210=2 \times 3 \times 5 \times 7\)
\(\displaystyle 55=5 \times 11\)
H.C.F. \(\displaystyle (210,55)=5\)
\(\displaystyle \therefore 5=210 \times 5+55 \mathrm{~m}\)
\(\displaystyle \Rightarrow \mathrm{m}=-19\)
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