Using the identity ( a + b + c )2 = a2 + b2 + c2 + 2(ab + bc + ac), we get x2+y+142= x22+y2+142+2 x2 #215;y+y #215;14+ x2 #215;14 #160; #160; #160; ...

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Standard VIII

Mathematics

Algebraic Identities

Expand -x2+y+...

Question

Expand ${(- \frac{x}{2} + y + \frac{1}{4})}^{2}$

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Solution

Using the identity ( a + b + c )² = a² + b² + c² + 2(ab + bc + ac), we get
${(- \frac{x}{2} + y + \frac{1}{4})}^{2} = {(- \frac{x}{2})}^{2} + y^{2} + {(\frac{1}{4})}^{2} + 2 [(- \frac{x}{2}) \times y + y \times \frac{1}{4} + (- \frac{x}{2}) \times \frac{1}{4}] = \frac{x^{2}}{4} + y^{2} + \frac{1}{16} + 2 [- \frac{x y}{2} + \frac{y}{4} - \frac{x}{8}] = \frac{x^{2}}{4} + y^{2} - x y + \frac{y}{2} - \frac{x}{4} + \frac{1}{16}$
Hence, ${(- \frac{x}{2} + y + \frac{1}{4})}^{2} = \frac{x^{2}}{4} + y^{2} - x y + \frac{y}{2} - \frac{x}{4} + \frac{1}{16}$ .

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Expand -x2+y+142

Using the identity ( a + b + c )2 = a2 + b2 + c2 + 2(ab + bc + ac), we get -x2+y+142=-x22+y2+142+2-x2×y+y×14+-x2×14 =x24+y2+116+2-xy2+y4-x8 =x24+y2-xy+y2-x4+116 Hence, -x2+y+142=x24+y2-xy+y2-x4+116.

Expand ${(- \frac{x}{2} + y + \frac{1}{4})}^{2}$