Question

If $\alpha ,\beta$ are the roots of the equation $2x^2-35x+2=0,$ then the value of $\sqrt{(2\alpha -35{)}^3(2\beta -35{)}^3}$ is

• A. 8.00
• B. 8
• C. 8.0

Answer 1

Answer: (A), (B), (C)

$2x^2-35x+2=0$
Let $y=2x-35$, where $x=\alpha ,\beta$
$x=\frac{y+35}{2}$
The equation whose roots are $2\alpha -35,2\beta -35$ is,
$f\left(\frac{x+35}{2}\right)=0$
$\Rightarrow 2{\left(\frac{x+35}{2}\right)}^2-35\left(\frac{x+35}{2}\right)+2=0$
$\Rightarrow x^2+35x+4=0$

Now, the product of the roots is,
$(2\alpha -35)(2\beta -35)=4$
$\therefore \sqrt{(2\alpha -35{)}^3(2\beta -35{)}^3}=\sqrt{4^3}=8$