For what value of k do the equations 2x^2 + k | vedclass

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(B) Let $\alpha$ be the common root of the two equations.Then,$2\alpha^2 + k\alpha - 5 = 0$ and $\alpha^2 - 3\alpha - 4 = 0$.Solving these using the c

Vedclass · Accepted Answer

$-3, -\frac{27}{4}$

Vedclass · Answer

(B) Let $\alpha$ be the common root of the two equations.Then,$2\alpha^2 + k\alpha - 5 = 0$ and $\alpha^2 - 3\alpha - 4 = 0$.Solving these using the cross-multiplication method:$\frac{\alpha^2}{-4k - 15} = \frac{\alpha}{-5 + 8} = \frac{1}{-6 - k}$.From this,$\alpha^2 = \frac{4k + 15}{k + 6}$ and $\alpha = \frac{-3}{k + 6}$.Since $\alpha^2 = (\alpha)^2$,we have $\left(\frac{-3}{k + 6}\right)^2 = \frac{4k + 15}{k + 6}$.$\frac{9}{(k + 6)^2} = \frac{4k + 15}{k + 6}$.$9 = (4k + 15)(k + 6)$.$9 = 4k^2 + 24k + 15k + 90$.$4k^2 + 39k + 81 = 0$.$(k + 3)(4k + 27) = 0$.Thus,$k = -3$ or $k = -\frac{27}{4}$.

For what value of $k$ do the equations $2x^2 + kx - 5 = 0$ and $x^2 - 3x - 4 = 0$ have a common root?

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