If the roots of the quadratic equation ax²+ bx + c = 0 are (k+1) / (k) & (k+2) / (k+1); then

[ a / (a+b+c) ]²equals

(A) k²

(B) (k+1)²

(C) (k+2)²

(C) k²(k+1)²

Question

If the roots of the quadratic equation ax2+ bx + c =
0 are (k+1) / (k) & (k+2) / (k+1); then
[ a / (a+b+c) ]2equals
(A) k2
(B) (k+1)2
(C) (k+2)2
(C) k2(k+1)2

Ankush Jain · Accepted Answer

Given, quadratic equation is: ax2+ bx + c = 0 and its roots are: $\frac{k + 1}{k} a n d \frac{k + 2}{k + 1}$ Now, sum of roots $= \frac{k + 1}{k} + \frac{k + 2}{k + 1} = \frac{- b}{a}$ $\Rightarrow \frac{b}{a} = - \frac{{(k + 1)}^{2} + k (k + 2)}{k (k + 1)} \Rightarrow \frac{b}{a} = - \frac{k^{2} + 1 + 2 k + k^{2} + 2 k}{k (k + 1)} = - \frac{2 k^{2} + 4 k + 1}{k (k + 1)} (1)$ Similarly, product of roots $= \frac{k + 1}{k} \times \frac{k + 2}{k + 1} = \frac{c}{a} \Rightarrow \frac{c}{a} = \frac{k + 2}{k} (2)$ So, ${(\frac{a}{a + b + c})}^{2} = {(\frac{1}{1 + \frac{b}{a} + \frac{c}{a}})}^{2} = {(\frac{1}{1 - \frac{2 k^{2} + 4 k + 1}{k (k + 1)} + \frac{k + 2}{k}})}^{2} [u \sin g (1) a n d (2)] = {(\frac{1}{\frac{k (k + 1) - 2 k^{2} - 4 k - 1 + (k + 1) (k + 2)}{k (k + 1)}})}^{2} = {(\frac{k (k + 1)}{k^{2} + k - 2 k^{2} - 4 k - 1 + k^{2} + 3 k + 2})}^{2} = {(\frac{k (k + 1)}{1})}^{2} = k^{2} (k + 1)^{2}$

If the roots of the quadratic equation ax2+ bx + c = 0 are (k+1) / (k) & (k+2) / (k+1); then [ a / (a+b+c) ]2equals (A) k2 (B) (k+1)2 (C) (k+2)2 (C) k2(k+1)2

If the roots of the quadratic equation ax²+ bx + c = 0 are (k+1) / (k) & (k+2) / (k+1); then

[ a / (a+b+c) ]²equals

(A) k²

(B) (k+1)²

(C) (k+2)²

(C) k²(k+1)²