Question

If the equation $2x^2-kx+x+8=0$ has real and equal roots, then $k=\underset{–}{}$

Answer 1

We have,

$2x^2-kx+x+8=0$

$\Rightarrow 2x^2-(k-1)x+8=0$

On comparing with general form of Quadratic Equation $a{x}^2+bx+c=0$

We get $a=2,b=-(k-1),c=8$

As roots are real and equal, so $D=0$

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

$\Rightarrow (-(k-1){)}^2-4\times 2\times 8=0$

$\Rightarrow k^2-2k+1-64=0$

$\Rightarrow k^2-2k-63=0$

$\Rightarrow k^2-9k+7k-63=0$

$\Rightarrow k(k-9)+7(k-9)=0$

$\Rightarrow (k-9)(k+7)=0$

$\Rightarrow k-9=0$ or $k+7=0$

$\Rightarrow k=9$ or $k=-7$

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