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Difference between the corresponding roots of $x^{2}+a x+b=0$ and $x^{2}+b x+a=0$ is same and $a \neq b$, then

1) $a+b+4=0$
2) $a+b-4=0$
3) $a-b-4=0$
4) $a-b+4=0$
in Mathematics by Diamond (75,350 points) | 8 views

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Answer

(1)

Explanation

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


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


Given $|\alpha-\beta|=|\gamma-\delta| \Rightarrow(\alpha-\beta)^{2}=(\gamma-\delta)^{2}$


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


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


$\Rightarrow a+b+4=0(\because a \neq b)$

by Diamond (75,350 points)

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