Question

The value of ${}^{\prime }{a}^{\prime }$ for which the quadratic equation $2x^2-x\left(a^2+8a-1\right)+a^2-4a=0$ has roots opposite signs, lie in the interval

• A. $1<a<5$
• B. $0<a<4$
• C. $-1<a<2$
• D. $2<a<6$

Answer 1

The correct option is B $0<a<4$

R.E.F image

Given equation is : $2x^2-x(a^2+8a-1)+a^2-4a=0$

Comparing it with the equation : $a{x}^2+bx+c=0$

we have $a=2$

$b=-(a^2+8a-1)$

$c=a^2-4a$

For the roots to have opposite signs.

product of roots $<0$

$\Rightarrow \frac{c}{a}<0$

$\Rightarrow \frac{a^2-4a}{2}<0$

$\Rightarrow a^2-4a<0$

$\Rightarrow a(a-4)<0$

$\Rightarrow (a-0)(a-4)<0$

$\therefore aE(0,4)$