Question

The values of 'a' and 'b' for which equation $( - \infty, - 2) \cup (2,\infty)$ have four real roots.

• A. – 6, – 4
• B. – 6, 5
• C. – 6, 4
• D. 6, – 4

Answer 1

Answer: (D)

6, – 4

Answer 2

Let for real roots are $ax^{2} + x(4a - b) + 4a + 2b + c = 0$ then equation is

$( x - \alpha ) ( x - \beta ) ( x - \gamma ) ( x - \delta ) = 0$ $x^{2} + bx + c = 0$ $+ \alpha \delta + \beta \delta + \alpha \gamma ) x ^ { 2 } - ( \alpha \beta \gamma + \beta \gamma \delta$ $r^{2}c = b^{2}q$

$r^{2}b = c^{2}q$

on comparing with $rb^{2} = cq^{2}$

$x^{2} - x - k = 0$

$2 \pm \sqrt{3}$

For real roots, A.M. of roots $3 \pm \sqrt{2}$G.M. of roots

$5 \pm \sqrt{2}$; $3 + 4i$

$x^{2} + px + q = 0p = 6,q = 25$

$p = 6,q = 1$⇒ $p = - 6,q = - 7$

$p = - 6,q = 25$ and $ax^{2} + bx + c = 0$

$\frac{b^{2}}{ac} + \frac{bc}{a^{2}} = \alpha$ = 1

Now, $\beta$

$x^{2} - 6x + a = 0$

$3\alpha + 2\beta = 16,$ = 6

$\alpha,\beta$

$lx^{2} + mx + n = 0$

$\alpha^{3}\beta$

$\alpha\beta^{3}$;$l^{4}x^{2} - nl(m^{2} - 2nl)x + n^{4} = 0$and $l^{4}x^{2} + nl(m^{2} - 2nl)x + n^{4} = 0$.