If a root of the equations x^2+px+q=0 and x^2 | vedclass

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(C) Let the common root be $y$. Then,$y^2+py+q=0$ and $y^2+\alpha y+\beta=0$.Subtracting the two equations: $(y^2+py+q) - (y^2+\alpha y+\beta) = 0$.$(

Vedclass · Accepted Answer

$\frac{q-\beta}{\alpha-p}$ or $\frac{p\beta-\alpha q}{q-\beta}$

Vedclass · Answer

(C) Let the common root be $y$. Then,$y^2+py+q=0$ and $y^2+\alpha y+\beta=0$.Subtracting the two equations: $(y^2+py+q) - (y^2+\alpha y+\beta) = 0$.$(p-\alpha)y + (q-\beta) = 0$.$y(p-\alpha) = \beta-q$.$y = \frac{\beta-q}{p-\alpha} = \frac{q-\beta}{\alpha-p}$.Alternatively,using the method of cross-multiplication for the system:$\frac{y^2}{p\beta-q\alpha} = \frac{y}{q-\beta} = \frac{1}{\alpha-p}$.From $\frac{y}{q-\beta} = \frac{1}{\alpha-p}$,we get $y = \frac{q-\beta}{\alpha-p}$.From $\frac{y^2}{p\beta-q\alpha} = \frac{y}{q-\beta}$,we get $y = \frac{p\beta-q\alpha}{q-\beta}$ (provided $q 
eq \beta$).

List-$I$	List-$II$
$(i) \ a \bar{a}$	$(A) -\frac{\pi}{3}$
$(ii) \ \arg \left(\frac{1}{\bar{a}}\right)$	$(B) -i \sqrt{3}$
$(iii) \ a-\bar{a}$	$(C) \frac{2i}{\sqrt{3}}$
$(iv) \ \operatorname{Im}\left(\frac{4}{3a}\right)$	$(D) 1$
	$(E) \frac{\pi}{3}$
	$(F) \frac{2}{\sqrt{3}}$

If a root of the equations $x^2+px+q=0$ and $x^2+\alpha x+\beta=0$ is common,then its value will be (where $p \neq \alpha$ and $q \neq \beta$).

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