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SSC MTS 9 Feb 2026 Shift 3 Paper

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Question : 13 of 150

Marks: +1, -0

If

a=4+\sqrt{7},b=4-\sqrt{7}

, then what is the value of

a^{2}+b^{2}

?

48
46
50
44

Solution:

Concept:

Use the identity

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

for conjugate surd pairs like

4+\sqrt{7}

and

4-\sqrt{7}

.

Explanation:

First, find

a+b

.

a+b=(4+\sqrt{7})+(4-\sqrt{7})=8

.

Next, find

ab

.

ab=(4+\sqrt{7})(4-\sqrt{7})=4^{2}-(\sqrt{7})^{2}=16-7=9

.

Now apply the identity:

a^{2}+b^{2}=8^{2}-2\times9=64-18=46

.

Shortcut: For

x\pm\sqrt{y}

,

a^{2}+b^{2}=2(x^{2}+y)

. With

x=4,y=7

, we get

2(16+7)=2\times23=46

.

Answer:

The correct value is 46, which matches option B.

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