Question

If the zeroes of the polynomial $64x^3-144x^2+92x-15$ are in A.P., then the difference between the largest and the smallest zeroes of the polynomial is

• A. $1$
• B. $\frac{7}{8}$
• C. $\frac{3}{4}$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $1$

Let $\alpha$ and $\beta$ be roots of the polynomial

Let $(\alpha +\beta ),\alpha ,(\alpha -\beta )$ be the three zeroes of the given polynomial $64x^3-144x^2+92x-15$

$\therefore (\alpha +\beta )+\alpha +(\alpha -\beta )=\frac{-(-144)}{64}=\frac{9}{4}$

$⟹3\alpha =\frac{9}{4}$

$⟹\alpha =\frac{3}{4}$

Also, $(\alpha +\beta )\alpha (\alpha -\beta )=\frac{-(-15)}{64}$

$⟹(\frac{3}{4}+\beta )\frac{3}{4}(\frac{3}{4}-\beta )=\frac{15}{64}$

$⟹(\frac{9}{16}-{\beta }^2)=\frac{5}{16}$

$⟹{\beta }^2=\frac{1}{4}$

$\therefore \beta =\pm \frac{1}{2}$

When $\alpha =\frac{3}{4}$ and $\beta =\frac{1}{2}$

So, the zeroes will be $\frac{5}{4},\frac{3}{4},\frac{1}{4}$

When $\alpha =\frac{3}{4}$ and $\beta =\frac{-1}{2}$

Then the zeroes will be $\frac{1}{4},\frac{3}{4},\frac{5}{4}$

$\therefore$ Difference between largest and smallest zeroes is $\frac{5}{4}-\frac{1}{4}=1$.

Hence, option A is correct.