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Q.
If the two equations $x^3-c x+d=0$ and $x^3-a x+b=0$ have one common root and the second equation has equal roots, then $2(b+d)=$
Complex Numbers and Quadratic Equations
Solution:
$\left(x-\frac{a}{2}\right)^2=\left(\frac{a^2}{4}-b\right)$
$\Rightarrow a^2 =4 b $..........(i)
$x =\frac{a}{2}$.............(ii)
$\alpha^2-c \alpha+d=0$
$\alpha^2-a \alpha+b=0$
$\alpha=\left(\frac{d-b}{c-a}\right)$ and from (ii) is $\frac{a}{2}$
$\frac{a}{2}=\frac{d-b}{c-a}=2 d-2 b=a c-a^2$
$\left\{\right.$ from (i) $\left.a^2=4 b\right\}$
$2(b+d)=a c$
