Exams › SSC CGL (Prelims) › General

Simplify: $\sqrt{17+12\sqrt{2}}$.

1. $4+\sqrt{2}$
2. $2+3\sqrt{2}$
3. $3+2\sqrt{2}$
4. $5+\sqrt{4}$

Correct answer: $3+2\sqrt{2}$

Solution

Write $\sqrt{17+12\sqrt2}=\sqrt{a}+\sqrt{b}$. Then $a+b=17$ and $2\sqrt{ab}=12\sqrt2$, so $ab=72$. The pair $a=9$, $b=8$ works, giving $\sqrt9+\sqrt8=3+2\sqrt2$.

Related SSC CGL (Prelims) General questions

• If $3x + 7 = 28$, find $x$.
• Simplify: \((15 + \sqrt{3}) + (15 - \sqrt{3}) - (1025 - 3)\)
• If $t=\sqrt{6}+3$, find $t^2-6\sqrt{6}$.
• If \(m + \frac{1}{m} = 6\), find the value of \(m^2 + \frac{1}{m^2}\).
• If $7x + 5 = 61$, then $x = ?$
• Find the y-intercept of \(5x + 3y = 15\).

⚔️ Practice SSC CGL (Prelims) General free + battle 1v1 →