Question

Suppose a and b are two roots of the equation$x^2-(\alpha -4)x+\alpha =0$. Find out the maximum possible value of$5ab-a^2-b^2.$

• A.
• B. 161/4
• C. 39
• D. 5

Answer 1

The correct option is B 161/4

Soln:

a and b are the roots of the equation

$a+b=(\alpha -4),ab=\alpha$

$5ab-a^2-b^2$

$5ab-(a^2+b^2)$

$5ab-\left[\right(a+b{)}^2-2ab]$

$5\alpha -\left[\right(\alpha -4{)}^2-2\alpha ]$

$5\alpha -[{\alpha }^2-10\alpha +16]$

$-{\alpha }^2+15\alpha -16=-({\alpha }^2+15\alpha -16)=-({\alpha }^2+16\alpha -\alpha -16)=-\left[\alpha \right(\alpha +16)-1(\alpha +16\left)\right]=-(\alpha +16)(\alpha -1)$

So,the maximum value$=\frac{D}{4a}=\frac{[{15}^2-4(-1\left)\right(-16\left)\right]}{4\times 1}=\frac{161}{4}.$Hence option (c).