If a, b are roots of quadratic equation (2-b) x2+(b2-3 b) x+b2=0 ; b ≠ 2 ; b ≠ 0 then

Question

If a, b are roots of quadratic equation (2-b) x2+(b2-3 b) x+b2=0 ; b ≠ 2 ; b ≠ 0 then (A) a, b, ab are in A.P. (B)  ab , a , b are in A.P. (C) 2 ab , 2 a , 2 b are in G.P. (D) (1/a), (1/b), (1/a b) are in H.P.

Tardigrade Admin · Accepted Answer

a b=(b2/2-b) ⇒ a=(b/2-b) ∴ a +1=(2/2- b ) ⇒( a +1) b =(2 b /2- b )=2 a ⇒ ab + b =2 a ∴ ab , a , b text are in A.P. ∴ 2 ab , 2 a , 2 b text are in G.P.

Q. If $a, b$ are roots of quadratic equation $(2 - b) x^{2} + (b^{2} - 3 b) x + b^{2} = 0; b \neq = 2; b \neq = 0$ then

a, b, ab are in A.P.

$ab, a, b$ are in A.P.

$2^{ab}, 2^{a}, 2^{b}$ are in G.P.

$\frac{1}{a}, \frac{1}{b}, \frac{1}{ab}$ are in H.P.

$ab = \frac{b ^{2}}{2 - b} \Rightarrow a = \frac{b}{2 - b}$
$∴ a + 1 = \frac{2}{2 - b} \Rightarrow (a + 1) b = \frac{2 b}{2 - b} = 2 a$
$\Rightarrow ab + b = 2 a$
$∴ ab, a, b are in A.P.$
$∴ 2^{ab}, 2^{a}, 2^{b} are in G.P.$

Q. If a,b are roots of quadratic equation (2−b)x2+(b2−3b)x+b2=0;b=2;b=0 then

a, b, ab are in A.P.

ab,a,b are in A.P.

2ab,2a,2b are in G.P.

a1​,b1​,ab1​ are in H.P.

ab=2−bb2​⇒a=2−bb​ ∴a+1=2−b2​⇒(a+1)b=2−b2b​=2a ⇒ab+b=2a ∴ab,a,b are in A.P. ∴2ab,2a,2b are in G.P.

Q. If $a, b$ are roots of quadratic equation $(2 - b) x^{2} + (b^{2} - 3 b) x + b^{2} = 0; b \neq = 2; b \neq = 0$ then

$ab, a, b$ are in A.P.

$2^{ab}, 2^{a}, 2^{b}$ are in G.P.

$\frac{1}{a}, \frac{1}{b}, \frac{1}{ab}$ are in H.P.

$ab = \frac{b ^{2}}{2 - b} \Rightarrow a = \frac{b}{2 - b}$
$∴ a + 1 = \frac{2}{2 - b} \Rightarrow (a + 1) b = \frac{2 b}{2 - b} = 2 a$
$\Rightarrow ab + b = 2 a$
$∴ ab, a, b are in A.P.$
$∴ 2^{ab}, 2^{a}, 2^{b} are in G.P.$