Question
If a = 2 b = -2 then find the value of
(i) $a^2+b^2$$\quad$ (ii) $a^2+a b+b^2$$\quad$ (iii) $a^2-b^2$
(i) $a^2+b^2$$\quad$ (ii) $a^2+a b+b^2$$\quad$ (iii) $a^2-b^2$
✓
Answer
(i) On substituting $a=2$ and $b=-2$ in $a^2+b^2$, we get
$a^2+b^2=(2)^2+(-2)^2$
= 4 + 4 = 8
(ii) On substituting $a=2$ and $b=-2$ in $a^2+a b+b^2$, we get
$a^2+a b+b^2=(2)^2+2 \times(-2)+(-2)^2$
= 4 - 4 + 4 = 4
(iii) On substituting $a=2$ and $b=-2$ in $a^2-b^2$, we get
$a^2-b^2=(2)^2-(-2)^2=4-4=0$
$a^2+b^2=(2)^2+(-2)^2$
= 4 + 4 = 8
(ii) On substituting $a=2$ and $b=-2$ in $a^2+a b+b^2$, we get
$a^2+a b+b^2=(2)^2+2 \times(-2)+(-2)^2$
= 4 - 4 + 4 = 4
(iii) On substituting $a=2$ and $b=-2$ in $a^2-b^2$, we get
$a^2-b^2=(2)^2-(-2)^2=4-4=0$
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