If $a + b = 1$ and $a - b = 7$ ; find $a^{2} + b^{2}$

22

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52

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32

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25

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Solution

The correct option is D $25$
Given,
$(a + b) = 1$
$(a - b) = 7$
Squaring the above two equation,we get,
$a^{2} + b^{2} + 2 a b = 1$ (i)
$a^{2} + b^{2} - 2 a b = 49$ (ii)
Adding the two equations, we get
$2 (a^{2} + b^{2}) = 50$
$=> (a^{2} + b^{2}) = \frac{50}{2}$
$=> (a^{2} + b^{2}) = 25$

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