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SSC CHSL 14 Nov 2025 Shift 1 Paper

Question : 72 of 100

Marks: +1, -0

a+b=\sqrt{17}

a-b=\sqrt{1}

\frac{(a^{2}+b^{2})}{(ab)}

Solution:

Concept:

Use the identities

(a+b)^2 = a^2 + b^2 + 2ab

and

(a-b)^2 = a^2 + b^2 - 2ab

to find

a^2 + b^2

and

ab

from the given sums and differences.

Explanation:

First, square the given values:

(a+b)^2 = (\sqrt{17})^2 = 17

(a-b)^2 = (\sqrt{1})^2 = 1

Now, add the two squared equations:

(a+b)^2 + (a-b)^2 = (a^2+b^2+2ab) + (a^2+b^2-2ab) = 2(a^2+b^2)

17 + 1 = 2(a^2+b^2)

→

18 = 2(a^2+b^2)

→

a^2+b^2 = 9

Next, subtract the second squared equation from the first:

(a+b)^2 - (a-b)^2 = (a^2+b^2+2ab) - (a^2+b^2-2ab) = 4ab

17 - 1 = 4ab

→

16 = 4ab

→

ab = 4

Finally, compute the required expression:

\frac{a^2+b^2}{ab} = \frac{9}{4}

Answer:

Option A:

\frac{9}{4}

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