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# Find the values of k for which the quadratic equation $(k + 4)x^{2} + ( k+1)x+1=0 $ has equal roots. Also find these roots.

**Given:**The quadratic equation (k + 4)x$^{2}$ + ( k+1)x+1=0 has equal roots. Also find these

**To do:**To find the value of k for which the given quadratic equation has equal roots.

**Solution:**

The given equation is,

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

On comparing it to the standard quadratic equation, $ax^{2} +bx+c=0$

$a=k+4,\ b=k+1\ and\ c=1$

For equal roots of any quadratic equation, its discriminant should be zero.

$D=0$

Or $b^{2} -4ac=0$

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

$\Rightarrow k^{2} +1+2k-4k-16=0$

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

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

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

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

If $k-5=0$

$\Rightarrow k=5$

If $k+3=0$

$\Rightarrow k=-3$

Therefore $k=5,\ -3$

If $k=5$, The equation becomes,

$( 5+4) x^{2} +( 5+1) x+1=0$

$\Rightarrow 9x^{2} +6x+1=0$

$\Rightarrow 9x^{2} +3x+3x+1=0$

$\Rightarrow 3x( 3x+1) +( 3x+1) =0$

$\Rightarrow ( 3x+1)( 3x+1) =0$

If $3x+1=0$

$\Rightarrow x=-\frac{1}{3}$

If $k=-3$, The equation becomes,

$( -3+4) x^{2} +( -3+1) x+1=0$

$x^{2} -2x+1=0$

$\Rightarrow x^{2} -x-x+1=0$

$\Rightarrow x( x-1) -( x-1) =0$

$\Rightarrow ( x-1)( x-1) =0$

If $x-1=0$

$\Rightarrow x=1$

For the value of k=5 or -3 the given quadratic equation have the equal roots.

And the equation has two equal roots $-\frac{1}{3}$ and 1.

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