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

- 25

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25

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12

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- 12

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Solution

The correct option is B $25$
Given, $a + b = 1$
Squaring both sides, we get
$(a + b)^{2} = 1^{2}$
$=> a^{2} + b^{2} + 2 a b = 1$ ------------(i)
Also $a - b = 7$
Squaring both sides, we get
$(a - b)^{2} = 7^{2}$
$=> a^{2} + b^{2} - 2 a b = 49$ ------------(ii)
Adding (i) and (ii), we get
$=> 2 (a^{2} + b^{2}) = 50$
$=> a^{2} + b^{2} = 25$

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If a+b=1 and a−b=7, then find a2+b2

The correct option is B 25Given, a+b=1Squaring both sides, we get (a+b)2=12=>a2+b2+2ab=1 ------------(i)Also a−b=7Squaring both sides, we get (a−b)2=72=>a2+b2−2ab=49 ------------(ii)Adding (i) and (ii), we get=>2(a2+b2)=50=>a2+b2=25

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