Question

State True or False:

The values of k for which the roots are real and equal of the following equation

$5x^2$ - 4x + 2 + k$(4x^2$ - 2x - 1) = 0 are $k=\frac{-6}{5},1$.

• A. True
• B. False

Answer 1

The correct option is B False

Given: $5x^2-4x+2+k(4x^2-2x-1)=0$

$\Rightarrow$ $5x^2-4x+2+4k{x}^2-2kx-k=0$

$\Rightarrow$ $(5+4k)x^2-(4+2k)x-k+2=0$

$\Rightarrow$ Here, $a=(5+4k),b=-(4+2k),c=-k+2$

$\Rightarrow$ It is given roots are real and equal.

$\therefore$ $b^2-4ac=0$

$\Rightarrow$ $[-(4+2k){]}^2-4(5+4k\left)\right(-k+2)=0$

$\Rightarrow$ $16+16k+4k^2-4(-5k+10-4k^2+8k)=0$

$\Rightarrow$ $16+16k+4k^2+20k-40+16k^2-32k=0$

$\Rightarrow$ $20k^2-4k-24=0$

$\Rightarrow$ $5k^2-k-6=0$

$\Rightarrow$ $5k^2-6k+5k-6=0$

$\Rightarrow$ $k(5k-6)+1(5k-6)=0$

$\Rightarrow$ $(5k-6)(k+1)=0$

$\Rightarrow$ $5k-6=0$ and $k+1=0$

$\Rightarrow$ $k=\frac{6}{5}$ and $k=-1$

$\Rightarrow$ We can see, values of $k$ given in question are incorrect.