If alpha, beta are the roots of the quadratic | vedclass

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Question

(B) Given equation is $x^{2}+a x+b=0, (b 
eq 0)$.Its roots are $\alpha$ and $\beta$. Then, sum of roots $\alpha+\beta = -a$ and product of roots $\al

Vedclass · Accepted Answer

$b x^{2}+a(b-1) x+(b-1)^{2}=0$

Vedclass · Answer

(B) Given equation is $x^{2}+a x+b=0, (b 
eq 0)$.Its roots are $\alpha$ and $\beta$. Then, sum of roots $\alpha+\beta = -a$ and product of roots $\alpha \beta = b$.Let the new roots be $S = \alpha-\frac{1}{\beta}$ and $T = \beta-\frac{1}{\alpha}$.Sum of new roots $= S+T = (\alpha+\beta) - (\frac{1}{\alpha} + \frac{1}{\beta}) = (\alpha+\beta) - \frac{\alpha+\beta}{\alpha \beta} = -a - \frac{-a}{b} = -a + \frac{a}{b} = \frac{a(1-b)}{b}$.Product of new roots $= ST = (\alpha-\frac{1}{\beta})(\beta-\frac{1}{\alpha}) = \alpha \beta - 1 - 1 + \frac{1}{\alpha \beta} = b - 2 + \frac{1}{b} = \frac{b^{2}-2b+1}{b} = \frac{(b-1)^{2}}{b}$.The required quadratic equation is $x^{2} - (	ext{sum of roots})x + (	ext{product of roots}) = 0$.$x^{2} - \frac{a(1-b)}{b}x + \frac{(b-1)^{2}}{b} = 0$.Multiplying by $b$, we get $b x^{2} - a(1-b)x + (b-1)^{2} = 0$.Since $-a(1-b) = a(b-1)$, the equation is $b x^{2} + a(b-1)x + (b-1)^{2} = 0$.

If $\alpha, \beta$ are the roots of the quadratic equation $x^{2}+a x+b=0, (b \neq 0),$ then the quadratic equation whose roots are $\alpha-\frac{1}{\beta}$ and $\beta-\frac{1}{\alpha}$ is:

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