Question

The values of k for which the roots are real and equal of the following equation
(2k + 1)$x^2$ + 2(k + 3)x + (k + 5) = 0 are $k=\frac{-5\pm \sqrt{41}}{2}$

• A. True
• B. False

Answer 1

The correct option is A True

$(2k+1)x^2+2(k+3)x+(k+5)=0$

$\Rightarrow$ Here, $a=2k+1,b=2(k+3),c=k+5$

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

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

$\Rightarrow$ $\left[2\right(k+3){]}^2-4(2k+1\left)\right(k+5)=0$

$\Rightarrow$ $4(k^2+6k+9)-4(2k^2+10k+k+5)=0$

$\Rightarrow$ $4k^2+24k+36-8k^2-40k-4k-20=0$

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

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

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

$\Rightarrow$ Here, $a=1,b=5,c=-4$

$\Rightarrow$ $k=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$=\frac{-5\pm \sqrt{(5{)}^2-4(1\left)\right(-4)}}{2\left(1\right)}$

$=\frac{-5\pm \sqrt{25+16}}{2}$

$\therefore$ $k=\frac{-5\pm \sqrt{41}}{2}$

$\therefore$ We can see value of $k$ given in question are correct.