Question

# Find the value of $$k$$ for which the equation $$x^2+k(2x+k-1)+2=0$$ has real and equal roots.

Solution

## Given, $$x^{2}+k(2x+k-1)+2=0$$Simplify above equation:$$x^{2}+2kx+(k^{2}-k+2)=0$$Compare given equation with the general form of quadratic equation, which $$ax^{2}+bx+c=0$$here, $$a=1, b=2k, c=(k^{2}-k+2)$$Find discriminant:$$D=b^{2}-4ac$$$$=(2k)^{2}-4\times 1\times (k^{2}-k+2)$$$$=4k^{2}-4k^{2}+4k-8$$$$=4k-8$$Since roots are real and equal (given)Put $$D=0$$$$4k-8=0$$$$k=2$$Hence, the value of $$k$$ is $$2$$.MathematicsRS AgarwalStandard X

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