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Q.
Let $p(x)4$ be a quadratic polynomial with constant term $1$. Suppose $p(x)$ when divided by $x - 1$ leaves remainder $2$ and when divided by $x + 1$ leaves remainder $4$. Then the sum of the roots of $p(x) = 0$ is
WBJEEWBJEE 2013Complex Numbers and Quadratic Equations
Solution:
Let $p(x)=a x^{2}+b x+c ...(i) $
Given, constant term ' $c$ '
$=1$ $\therefore p(x)=a x^{2}+b x+1 ...(ii) $
Now, by given condition, $p(1)=2 $ (remainder)
$\Rightarrow a+b+1=2$
$\Rightarrow a+b=1 ...(iii) $
and $p(-1)=4 $ (remainder)
$\Rightarrow a-b+1=4$
$\Rightarrow a-b=3...(iii) $
On adding Eqs. (iii) and (iv), we get
$2 a=4 $
$\Rightarrow a=2$ form eys (iii) $b=-1$
On putting the values of a and bin Eq (ii), we get
$ p(x)=2 x^{2}-x+1=0 $
$ \therefore $ Sum of the roots $=-\frac{(\text { Coefficient of } x)}{\left(\text { Coefficient of } x^{2}\right)}=\frac{-(-1)}{2} $
$=\frac{1}{2}$