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4.(vi)  Expand each of the following, using suitable identities:

     (vi)    \left [\frac{1}{4}a - \frac{1}{2}b + 1\right ]^2

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Given is  \left [\frac{1}{4}a - \frac{1}{2}b + 1\right ]^2

We will Use identity  

(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2zx

Here, x =\frac{a}{4} , y = -\frac{b}{2} \ \ and \ \ z = 1

Therefore,  

\left [\frac{1}{4}a - \frac{1}{2}b + 1\right ]^2=\left(\frac{a}{4} \right )^2+\left(-\frac{b}{2} \right )^2+(1)^2+2.\left(\frac{a}{4} \right ). \left(-\frac{b}{2} \right )+2. \left(-\frac{b}{2} \right ).1+2.1.\left(\frac{a}{4} \right )

                               = \frac{a^2}{16}+\frac{b^2}{4}+1-\frac{ab}{4}-b+\frac{a}{2}

                                

Posted by

Gautam harsolia

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