Let a, b, c, p, q be real numbers. Suppose al | vedclass

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(B) For the first equation $x^2+2px+q=0$,the roots are $\alpha, \beta$. Thus,$\alpha+\beta = -2p$ and $\alpha\beta = q$. The discriminant is $D_1 = 4p

Vedclass · Accepted Answer

$STATEMENT-1$ is True,$STATEMENT-2$ is True; $STATEMENT-2$ is $NOT$ a correct explanation for $STATEMENT-1$.

Vedclass · Answer

(B) For the first equation $x^2+2px+q=0$,the roots are $\alpha, \beta$. Thus,$\alpha+\beta = -2p$ and $\alpha\beta = q$. The discriminant is $D_1 = 4p^2-4q = 4(p^2-q)$.For the second equation $ax^2+2bx+c=0$,the roots are $\alpha, \frac{1}{\beta}$. Thus,$\alpha+\frac{1}{\beta} = -\frac{2b}{a}$ and $\alpha\cdot\frac{1}{\beta} = \frac{c}{a}$. The discriminant is $D_2 = 4b^2-4ac = 4(b^2-ac)$.Since $\alpha$ is a common root,we have $\alpha^2+2p\alpha+q=0$ and $a\alpha^2+2b\alpha+c=0$. Eliminating $\alpha^2$,we get $(2ap-2b)\alpha + (aq-c) = 0$. If $b=ap$ and $c=aq$,then the equations are identical,which contradicts $\beta^2 
eq 1$. Thus $b 
eq ap$ or $c 
eq aq$.Since $\alpha$ is real,$p^2-q \geq 0$ and $b^2-ac \geq 0$ (as $\alpha$ is a root of both),their product is $\geq 0$. Thus $STATEMENT-1$ is True.$STATEMENT-2$ is also True as shown by the contradiction of identical equations,but it is not the direct explanation for the inequality product. Hence,option $B$ is correct.

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