Question

For what value of k, equation: 2x^2+kx-5 = 0 & x^2 - 3x - 4 = 0 may have one root common.

Answer 1

Let a be the common root of the given equations, then

The Equation will be 2a^2 + ka - 5 = 0  ------ (1)

The Equation will be a^2 - 3a - 4 = 0  ------- (2)


On solving (1) & (2) * 2, we get

= > 2a^2 + ka = 5

= > 2a^2 - 6a = 8
        
   -----------------------

                ka + 6a = -3

                a(k + 6) = -3

                 a = -3/(k + 6).


Substitute a = -3/(k + 6) in (1), we get

$= \ \textgreater \ 2(\frac{-3}{k + 6} )^2 + k( -\frac{3}{k + 6} ) - 5 = 0$

$= \ \textgreater \ \frac{18}{k^2 + 12k + 36} - \frac{3k}{k + 6} - 5 = 0$

LCM of k^2 + 12k + 36, k + 6 = (k + 6)^2.

$= \ \textgreater \ \frac{18}{k^2 + 12k + 36} * (k + 6)^2 - \frac{3k}{k + 6} * (k + 6)^2 - 5(k + 6)^2 = 0$

$= \ \textgreater \ 18 - 3k(k + 6) - 5(k + 6)^2 = 0$

$= \ \textgreater \ 18 - 3k^2 - 18k - 5(k^2 + 36 + 12k) = 0$

$= \ \textgreater \ 18 - 3k^2 - 18k - 5k^2 - 180 - 60k = 0$

$= \ \textgreater \ -8k^2 - 78k - 162 = 0$

$= \ \textgreater \ -2(4k^2 + 39k + 81) = 0$

$= \ \textgreater \ 4k^2 + 39k + 81 = 0$

$= \ \textgreater \ 4k^2 + 12k + 27k + 81 = 0$

$= \ \textgreater \ 4k(k + 3) + 27(k + 3) = 0$

$= \ \textgreater \ (4k + 27)(k + 3) = 0$

$= \ \textgreater \ k = \frac{-27}{4} , -3$



Therefore, the required values of k are : (-3), (-27/4).


Hope this helps!