Suppose a, b ∈ R , a ≠ 0 and 2 a+b ≠ 0. A root of the equation (a+b)(a x+b)(a-b x)=(a2 x-b2)(a+b x) is

Question

Suppose a, b ∈ R , a ≠ 0 and 2 a+b ≠ 0. A root of the equation (a+b)(a x+b)(a-b x)=(a2 x-b2)(a+b x) is (A) (a+2 b/2 a+b) (B) (2 a+b/a+2 b) (C) -(a-2 b/a+2 b) (D) -(a+2 b/2 a+b)

Tardigrade Admin · Accepted Answer

We can write the quation as [a2 b+(a+b) a b] x2+[(a3-b3)-(a+b)(a2-b2)] x -a b2-a b(a+b)=0 ⇒(2 a+b) x2-(a-b) x-(a+2 b)=0 Since the sum of coefficients is 0 one of the roots is 1 and the other root is -(a+2 b/2 a+b).

Q. Suppose $a, b \in R, a \neq = 0$ and $2 a + b \neq = 0$ . A root of the equation $(a + b) (a x + b) (a - b x) = (a^{2} x - b^{2}) (a + b x)$ is

$\frac{a + 2 b}{2 a + b}$

$\frac{2 a + b}{a + 2 b}$

$- \frac{a - 2 b}{a + 2 b}$

$- \frac{a + 2 b}{2 a + b}$

We can write the quation as
$[a^{2} b + (a + b) ab] x^{2} + [(a^{3} - b^{3}) - (a + b) (a^{2} - b^{2})] x - a b^{2} - ab (a + b) = 0$
$\Rightarrow (2 a + b) x^{2} - (a - b) x - (a + 2 b) = 0$
Since the sum of coefficients is 0 one of the roots is 1 and the other root is $- \frac{a + 2 b}{2 a + b}$ .

Q. Suppose a,b∈R,a=0 and 2a+b=0. A root of the equation (a+b)(ax+b)(a−bx)=(a2x−b2)(a+bx) is

2a+ba+2b​

a+2b2a+b​

−a+2ba−2b​

−2a+ba+2b​

We can write the quation as [a2b+(a+b)ab]x2+[(a3−b3)−(a+b)(a2−b2)]x−ab2−ab(a+b)=0 ⇒(2a+b)x2−(a−b)x−(a+2b)=0 Since the sum of coefficients is 0 one of the roots is 1 and the other root is −2a+ba+2b​.

Q. Suppose $a, b \in R, a \neq = 0$ and $2 a + b \neq = 0$ . A root of the equation $(a + b) (a x + b) (a - b x) = (a^{2} x - b^{2}) (a + b x)$ is

$\frac{a + 2 b}{2 a + b}$

$\frac{2 a + b}{a + 2 b}$

$- \frac{a - 2 b}{a + 2 b}$

$- \frac{a + 2 b}{2 a + b}$