First slide

Theory of equations

Question

If the equations $2 x^{2} + k x - 5 = 0$ and $x^{2} - 3 x - 4 = 0$ have one root in common, then $k =$

Moderate

A

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

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

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

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

Solution

Let a be the common root of the two equations
Then,
$2 α^{2} + k α - 5 = 0 and, α^{2} - 3 α - 4 = 0$
Solving these two equations, we get
$\frac{α^{2}}{- 4 k - 15} = \frac{α}{- 5 + 8} = \frac{1}{- 6 - k}$
$\begin{array}{l} \Rightarrow α^{2} = \frac{4 k + 15}{k + 6} and α = \frac{- 3}{k + 6} \\ \Rightarrow {(\frac{- 3}{k + 6})}^{2} = \frac{4 k + 15}{k + 6} [∵ α^{2} = (α)^{2}] \\ \Rightarrow (4 k + 15) (k + 6) = 9 \\ \Rightarrow 4 k^{2} + 39 k + 81 = 0 \Rightarrow k = - 3 or, k = - \frac{27}{4} \end{array}$

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