Question

using suitable identity expand (1/2a-1/2b+1)^2

Answer 1

using (a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ac
=(1/2a)^2+(-1/2b)^2+1^2+2×1/2a×-1/2b+2×-1/2b×1+2×1/2a×1
=1/4a^2+1/4b^2+1-1/2ab-1+1

Answer 2

The expanded form is $(\frac{1}{2}a-\frac{1}{2}b+1)^2=\frac{1}{4}a^2+\frac{1}{4}b^2+1-\frac{1}{2}ab-b+a$

Step-by-step explanation:

Given : Expression $(\frac{1}{2}a-\frac{1}{2}b+1)^2$

To find : Using suitable identity and expand?

Solution :

Using algebraic identity,

$(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2xz$

Here, $x=\frac{1}{2}a,\ y=\frac{1}{2}b,\ z=1$

$(\frac{1}{2}a-\frac{1}{2}b+1)^2=(\frac{1}{2}a)^2+(-\frac{1}{2}b)^2+1^2+2\times (\frac{1}{2}a)\times (-\frac{1}{2}b)+2\times (-\frac{1}{2}b)\times 1+2\times (\frac{1}{2}a)\times 1$

$(\frac{1}{2}a-\frac{1}{2}b+1)^2=\frac{1}{4}a^2+\frac{1}{4}b^2+1-\frac{1}{2}ab-b+a$

Therefore, the expanded form is $(\frac{1}{2}a-\frac{1}{2}b+1)^2=\frac{1}{4}a^2+\frac{1}{4}b^2+1-\frac{1}{2}ab-b+a$

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