Question

If $a>2,$ then the roots of the equation $\left(2-a\right)x^2+3ax-1=0$ are

• A. one positive and one negative
• B. both negative
• C. both positive
• D. both imaginary

Answer 1

Answer: (C)

Let, $f\left(x\right)=x^2-\frac{3a}{a-2}x+\frac{1}{a-2}=0$
$D=\frac{9a^2}{{\left(a-2\right)}^2}-4\frac{1}{\left(a-2\right)}=\frac{9a^2-4a+8}{{\left(a-2\right)}^2}$
$=\frac{8a^2+{\left(a-2\right)}^2+4}{{\left(a-2\right)}^2}>0$
$f\left(0\right)=\frac{1}{a-2}>0$ and
$\frac{-B}{2A}=\frac{3a}{2\left(a-2\right)}>0$
Since, $D>0,f\left(0\right)>0,\frac{-B}{2A}>0$
Hence, both roots of the given equation are positive.