If 3x^2 - 7x + 2 = 0 and 15x^2 - 11x + a = 0 | vedclass

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(C) Given equations are $3x^2 - 7x + 2 = 0$ $(i)$ and $15x^2 - 11x + a = 0$ (ii). Solving $(i)$: $3x^2 - 6x - x + 2 = 0 \implies 3x(x - 2) - 1(x - 2)

Vedclass · Accepted Answer

$\frac{2}{15}$

Vedclass · Answer

(C) Given equations are $3x^2 - 7x + 2 = 0$ $(i)$ and $15x^2 - 11x + a = 0$ (ii). Solving $(i)$: $3x^2 - 6x - x + 2 = 0 \implies 3x(x - 2) - 1(x - 2) = 0 \implies (3x - 1)(x - 2) = 0$. So,the roots of $(i)$ are $x = \frac{1}{3}$ and $x = 2$. Case $1$: If $x = 2$ is a common root,then $15(2)^2 - 11(2) + a = 0 \implies 60 - 22 + a = 0 \implies a = -38$. Since $a > 0$,this is rejected. Case $2$: If $x = \frac{1}{3}$ is a common root,then $15(\frac{1}{3})^2 - 11(\frac{1}{3}) + a = 0 \implies 15(\frac{1}{9}) - \frac{11}{3} + a = 0 \implies \frac{5}{3} - \frac{11}{3} + a = 0 \implies -2 + a = 0 \implies a = 2$. Now,for the equation $15x^2 - ax + 7 = 0$,substituting $a = 2$,we get $15x^2 - 2x + 7 = 0$. The sum of the roots is given by $-\frac{b}{A} = -(\frac{-2}{15}) = \frac{2}{15}$.

If $3x^2 - 7x + 2 = 0$ and $15x^2 - 11x + a = 0$ have a common root and $a$ is a positive real number,then the sum of the roots of the equation $15x^2 - ax + 7 = 0$ is:

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