Question

If the roots of the equation $3x^2+2(a^2+1)x+(a^2-674x+2013)=0$ are of opposite sign, then the value of ${}^{\prime }{a}^{\prime }$ in the quadratic equation is

• A. $[3,671]$
• B. $(3,671]$
• C. $[3,671)$
• D. $(3,671)$

Answer 1

The correct option is D $(3,671)$

$Givene{q}^n$

$\Rightarrow 3x^2+2(a^2+1)x+(a^2-674x+2013)=0$

Condition = (1): roots should be real

$\Rightarrow b^2-4ac$

Condition (2) : Product of roots must be -ve

or $(\frac{c}{a}=0)$

$\therefore C1:\left(2\right(a^2+1){)}^2>4.3(a^{2}674a+2013)$

$\Rightarrow (a^2+1{)}^1>3(a-3\left)\right(a-671)$

$C2:\left(\frac{a^2-674a+2013}{3}\right)<0$

$\Rightarrow (a-3)(a-671)<0$

$\therefore aϵ(3,671)$

Option D is correct