Difference between the corresponding roots of x^2+ax+b=0 and x^2+bx+a=0 is the same, and a…

Question

Difference between the corresponding roots of $x^2+ax+b=0$ and $x^2+bx+a=0$ is the same, and $a
e b$, then

Accepted Answer

Let $$\alpha ,\beta $$ and $$\gamma ,\delta $$ be the roots of the equations ${x^2} + ax + b = 0$

and ${x^2} + bx + a = 0$ respectively.

$$\therefore$$ $$\alpha  + \beta  =  - a,\alpha \beta  = b$$

and $$\gamma  + \delta  =  - b,\gamma \delta  = a.$$

Given $$\left| {\alpha  - \beta } \right| = \left| {\gamma  - \delta } \right|$$

$$ \Rightarrow {\left( {\alpha  - \beta } \right)^2} = {\left( {\gamma  - \delta } \right)^2}$$

$$ \Rightarrow {\left( {\alpha  + \beta } \right)^2} - 4\alpha \beta  = {\left( {\gamma  + \delta } \right)^2} - 4\gamma \delta $$

$$ \Rightarrow {a^2} - 4b = {b^2} - 4a$$

$$ \Rightarrow \left( {{a^2} - {b^2}} \right) + 4\left( {a - b} \right) = 0$$

$$ \Rightarrow a + b + 4 = 0$$

( as $$a \ne b$$ )

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