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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.
Updated on: 10-Oct-2022
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