Question

If the roots of
$x^{4}-10 x^{3}+37 x^{2}-60 x+36=0$ are
$\alpha, \alpha, \beta, \beta(\alpha<\beta)$, then $2 \alpha+3 \beta-2 \alpha \beta=$

• A. 1
• B. 0
• C. -1
• D. 4

Answer 1

Answer: (A)

We have,
$x^{4}-10 x^{3}+37 x^{2}-60 x+36=0$
since, $\alpha, \alpha, \beta, \beta$ are the roots of the above equation.
$\therefore \alpha+\alpha+\beta+\beta=\frac{-(-10)}{1}=10$
$\Rightarrow \alpha+\beta=5$ ...(i)
and $(\alpha)(\alpha)(\beta)(\beta)=\frac{36}{1}$
$\Rightarrow \alpha^{2} \beta^{2}=36$
$\Rightarrow \alpha \beta=6$ ...(ii)
From Eqs. (i) and (ii), we have
$\alpha(5-\alpha)=6$
$\Rightarrow \alpha^{2}-5 \alpha+6=0$
$\Rightarrow \alpha=2,3$
$\Rightarrow \beta=3,2$
Since $\alpha<\beta,$ hence $\alpha =2,\, \beta=3$
$\therefore 2 \alpha+3 \beta-2 \alpha \beta$
$=2 \times 2+3 \times 3-2 \times 2 \times 3$
$=4+9-12=1$