Question

Solving the equation
$\sqrt{({2x}^2+5x-2)}-\sqrt{{2x}^2+5x-9}=1$
we get two solutions of x ,let them be $k,m$.Find $-(k\ast m)$ ?

Answer 1

Let u= $\sqrt{({2x}^2+5x-2)}$
and v = $\sqrt{{2x}^2+5x-9}$
$\therefore u^2={2x}^2+5x-2and{v}^2={2x}^2+5x-9$
So, the given equation reduces to $u-v=1$ ....(1)
Now, $u^2-v^2=7$
$\Rightarrow (u+v)(u-v)=7$
$\Rightarrow u+v=7(\because u-v=1)$ ....(2)
Solving (1) and (2),we get
$u=4,v=3$
$\therefore \sqrt{({2x}^2+5x-2)}=4$
$\Rightarrow {2x}^2+5x-18=0$
$\Rightarrow x_1=2;x_2=-\frac{9}{2}$
Both roots satisfies the original equation .
Hence $x_1=2$ and $x_2=-\frac{9}{2}$ are the roots of the original equation.
Required product $=-2\times (-\frac{9}{2})=9$