Question

The values of $k$, so that the equations $2x^2+kx-5=0$ and $x^2-3x-4=0$ have one root in common, are

• A. $3,\frac{27}{2}$
• B. $9,\frac{27}{4}$
• C. $-3,\frac{-27}{4}$
• D. $-3,\frac{4}{27}$

Answer 1

The correct option is C $-3,\frac{-27}{4}$

$2x^2+kx-5=0$
$x^2-3x-4=0$----- Given
$(c_1{a}_2-c_2{a}_1{)}^2=(b_1{c}_2-b_2{c}_1\left)\right(a_1{b}_2-a_2{b}_1)$
$\Rightarrow [-5\times 1+4\times 2{]}^2=[K(-4)+3(-5)\left]\right[2(-3)-\left(1\right)\left(K\right)]$
$\therefore 3^2=[-4K-15][-6-K]$
$\therefore 3^2=(4K+15)\left(65K\right)$
$\therefore 3^2=(4K+15)(64K_{)}$
$\therefore 9=24K+4K2+90+15K$
$\therefore 0=4K^2+39K+81$
$\therefore K=-3,-\frac{27}{4}$

Alternate solution

$x^2-3x-4=0$
$x^2-4x+x-4=0$
$\therefore (x-4)(x+1)=0$
$\therefore x=4,-1$
By keeping $x=4$
we get, $2(4{)}^2+k\times 4-5=0$
$4k=-27$

$\therefore k=\frac{-27}{4}$
by keeping $x=-1$
we get, $\therefore 2(-1{)}^2+k(-1)-5=0$
$\therefore k=-3$.