Question

If $m(a{x}^2+2bx+c)+p{x}^2+2qx+r$ can be expressed in the form of $n(x+k{)}^2$, then the value of $(ak-b)(qk-r)$ is

• A. $(pk-q)(bk-c)$
• B. $\frac{(pk-q)}{(bk-c)}$
• C. $-\frac{(pk-q)}{(bk-c)}$
• D. $-(pk-q)(bk-c)$

Answer 1

The correct option is A $(pk-q)(bk-c)$

$m(a{x}^2+2bx+c)+p{x}^2+2qx+r=n(x+k{)}^2$
Equating the coefficients of same power of $x$, we get
$ma+p=n...\left(1\right)$
$mb+q=nk....\left(2\right)$
$mc+r=n{k}^2...\left(3\right)$

From equation $\left(1\right)$ and $\left(2\right)$,
$mb+q=k(ma+p)$
$\Rightarrow m(ak-b)+pk-q=0\Rightarrow m=-\frac{pk-q}{ak-b}....\left(4\right)$
From equation $\left(2\right)$ and $\left(3\right)$,
$mc+r=k(mb+q)$
$\Rightarrow m(bk-c)+qk-r=0\Rightarrow m=-\frac{qk-r}{bk-c}....\left(5\right)$

From equation $\left(4\right)$ and $\left(5\right)$,
$\frac{pk-q}{ak-b}=\frac{qk-r}{bk-c}$
$\Rightarrow (ak-b)(qk-r)=(pk-q)(bk-c)$