Question

7. If The Equations X2 - Ax + B = 0 And X2 – Ex + F = 0 Have One Root In Common And If The Seaequation Has Equal Roots, Then Prove That Ae = 2(B + F).​

Answer 1

Answer:

$\bold{2(f+b)=ae}$

Step-by-step explanation:

Given equations are

$x^2-ax+b=0\:and\:x^2-ex+f=0$

Let the common root be $\alpha$

Then

${\alpha}^2-a{\alpha}+b=0$

${\alpha}^2-e{\alpha}+f=0$

subtracting we get

$-a{\alpha}+e{\alpha}+b-f=0$

$(e-a){\alpha}+b-f=0$

$(e-a){\alpha}=f-b$

$\alpha=\frac{f-b}{e-a}$......(1)

since the equation $x^2-ex+f=0$ has equal roots,

$b^2-4ac=0$

$\implies\:(-e)^2-4f=0$

$\implies\:e^2=4f$......(2)

since the equation $x^2-ex+f=0$ has equal roots,

sum of the roots $=\frac{e}{1}$

$\implies\:\alpha+\alpha=\frac{e}{1}$

$\implies\:2\alpha=e$

$\implies\:\alpha=\frac{e}{2}$.....(3)

using (3) in (1) we get

$\frac{e}{2}=\frac{f-b}{e-a}$

$\implies\:e(e-a)=2(f-b)$

$\implies\:4f-ae=2f-2b)$

Rearranging terms we get

$\implies\:4f-2f+2b=ae$

$\implies\:2f+2b=ae$

$\implies\:2(f+b)=ae$