If the quadratic equations 3x^2 - 7x + 2 = 0 | vedclass

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(A) Let $\alpha$ be the common root of the equations $3x^2 - 7x + 2 = 0$ and $kx^2 + 7x - 3 = 0$. Then,$3\alpha^2 - 7\alpha + 2 = 0$ $\dots(i)$ and $k

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(A) Let $\alpha$ be the common root of the equations $3x^2 - 7x + 2 = 0$ and $kx^2 + 7x - 3 = 0$. Then,$3\alpha^2 - 7\alpha + 2 = 0$ $\dots(i)$ and $k\alpha^2 + 7\alpha - 3 = 0$ $\dots(ii)$. Adding $(i)$ and $(ii)$,we get $(k+3)\alpha^2 - 1 = 0$,which implies $\alpha^2 = \frac{1}{k+3}$. From $(i)$,$3\alpha^2 + 2 = 7\alpha$,so $3(\frac{1}{k+3}) + 2 = 7\alpha$,which gives $\alpha = \frac{2k+9}{7(k+3)}$. Substituting $\alpha^2$ into the equation,we have $\frac{1}{k+3} = \left(\frac{2k+9}{7(k+3)}\right)^2$. $49(k+3) = (2k+9)^2 = 4k^2 + 36k + 81$. $4k^2 - 13k - 66 = 0$. $(k-6)(4k+11) = 0$. Since $k$ is positive,$k = 6$.

If the quadratic equations $3x^2 - 7x + 2 = 0$ and $kx^2 + 7x - 3 = 0$ have a common root,then the positive value of $k$ is

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