If the roots of the quadratic equation 3x2+2x+a2-a=0 in x are of opposite signs, then a lies in the interval

Question

If the roots of the quadratic equation 3x2+2x+a2-a=0 in x are of opposite signs, then a lies in the interval (A) (-∞, -2) (B)  (-∞ ,0)  (C)  (-1,0)  (D)  (0,1)

Tardigrade Admin · Accepted Answer

3x2+2x+a2-a=0 The condition for a quadratic equation have both roots are an opposite sign is constant term <0 ⇒ (a2-a)<0 ⇒ a(a-1)<0 ⇒ a∈ (0,1)

Q. If the roots of the quadratic equation $3 x^{2} + 2 x + a^{2} - a = 0$ in $x$ are of opposite signs, then a lies in the interval

$(- \infty, - 2)$

$(- \infty, 0)$

$(- 1, 0)$

$(0, 1)$

$(1, 3)$

$3 x^{2} + 2 x + a^{2} - a = 0$
The condition for a quadratic equation have both roots are an opposite sign is constant term $< 0$

$\Rightarrow$ $(a^{2} - a) < 0$
$\Rightarrow$ $a (a - 1) < 0$
$\Rightarrow$ $a \in (0, 1)$

Q. If the roots of the quadratic equation 3x2+2x+a2−a=0 in x are of opposite signs, then a lies in the interval

(−∞,−2)

(−∞,0)

(−1,0)

(0,1)

(1,3)

3x2+2x+a2−a=0 The condition for a quadratic equation have both roots are an opposite sign is constant term <0 ⇒ (a2−a)<0 ⇒ a(a−1)<0 ⇒ a∈(0,1)