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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 $$2$$.

Mathematics
RS Agarwal
Standard X

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