Let x1, x2 be the roots of x2 - 3x + a = 0 and x3, x4 be the roots of x2 - 12x + b = 0. If x1

Question

Let x1, x2 be the roots of x2 - 3x + a = 0 and x3, x4 be the roots of x2 - 12x + b = 0. If x1 < x2 < x3 < x4 and x 1 , x2, x3, x4 are in G.P. then ab equals (A) (24/5) (B) 64 (C) 16 (D) 8

Tardigrade Admin · Accepted Answer

x1 + x2 = 3; x1.x2 = a x3 + x4 = 12; x3.x4 = b Let r be the common ratio of G.P., then (x1(1+r)/x1r2(1+r)) = (3/12) ⇒ r = ±2 Take r = 2 (∵ G.P. is increasing) ∵ x1 +x2 = 3 ⇒ x1(1+r) = 3 ⇒ x1 = 1 ∴ ab = x1 x2 x3 x4 = 1.2.4.8 = 64

Q. Let $x_{1}, x_{2}$ be the roots of $x^{2} - 3 x + a = 0$ and $x_{3}, x_{4}$ be the roots of $x_{2} - 12 x + b = 0$ . If $x_{1} < x_{2} < x_{3} < x_{4}$ and $x_{1}, x_{2}, x_{3}, x_{4}$ are in G.P. then ab equals

$\frac{24}{5}$

64

16

8

Q. Let x1​,x2​ be the roots of x2−3x+a=0 and x3​,x4​ be the roots of x2​−12x+b=0. If x1​<x2​<x3​<x4​ and x1​,x2​,x3​,x4​ are in G.P. then ab equals

524​

64

16

8

x1​+x2​=3;x1​.x2​=a x3​+x4​=12;x3​.x4​=b Let r be the common ratio of G.P., then x1​r2(1+r)x1​(1+r)​=123​⇒r=±2 Take r=2 (∵ G.P. is increasing) ∵x1​+x2​=3⇒x1​(1+r)=3⇒x1​=1 ∴ab=x1​x2​x3​x4​=1.2.4.8=64

Q. Let $x_{1}, x_{2}$ be the roots of $x^{2} - 3 x + a = 0$ and $x_{3}, x_{4}$ be the roots of $x_{2} - 12 x + b = 0$ . If $x_{1} < x_{2} < x_{3} < x_{4}$ and $x_{1}, x_{2}, x_{3}, x_{4}$ are in G.P. then ab equals

$\frac{24}{5}$