Question

Find the values of $k$ for the following quadratic equation, so that they have two real and equal roots:

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

• A. $k=3,5$
• B. $k=3,-5$
• C. $k=-3,-5$
• D. $k=-3,5$

Answer 1

Answer: (D)

$k=-3,5$

Given quadratic equation is $4x^2-2(k+1)x+(k+4)=0$

Now given that the roots are equal and zero.

For a quadratic equation $a{x}^2+bx+c=0$ roots are real and equal, if

$D=b^2-4ac$

Here, $a=4,b=-2(k+1),c=(k+4)$

Therefore, $D=b^2-4ac$

$\Rightarrow [-2(k+1){]}^2-4\times 4\times (k+4)=0$

$\Rightarrow 4(k^2+2k+1)-16k-64=0$

$\Rightarrow k2+2k+1-4k-16=0$

$\Rightarrow k^2-2k-15=0$

$\Rightarrow k^2-5k+3k-15=0$

$\Rightarrow k(k-5)+3(k-5)=0$

$\Rightarrow (k-5)(k+3)=0$

$\Rightarrow k=5$ and $k=-3$