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the question is the expression $kx^2 +(k+1)x +2$ will be a perfect square of a linear polynomial for what values of k .

I am unable to understand the concept used in this question for finding the possible values for k.

please someone explain.

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  • $\begingroup$ $(mx+b)^2=m^2x^2 + 2mbx +b^2$ 2 so $b = \sqrt2$ and $k=m^2$ so $m=\sqrt k $ and $k+1 =2mb=2\sqrt {2k}$ . Solve for $k $. $(k+1)^2 = 8k $ so $k^2 -6k +1 =0$. So $k = 3 \pm \sqrt {32}/2=3\pm 2\sqrt2$. $\endgroup$ – fleablood Jul 22 '16 at 8:13
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$$k\left(x^2+\dfrac{(k+1)x}{2k}+\left(\dfrac{k+1}{2k}\right)^2\right)+2-\dfrac{(k+1)^2}{4k}$$

So, we need $$2-\dfrac{(k+1)^2}{4k}=0\iff k=?$$

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  • $\begingroup$ $k=3+\sqrt2$ and $k=3-\sqrt2$, but why did you put that equals to zero? $\endgroup$ – danny Jul 22 '16 at 5:51
  • $\begingroup$ @danny. First, you complete the square and what is left must be $0$; so $k$. $\endgroup$ – Claude Leibovici Jul 22 '16 at 5:55
  • $\begingroup$ @ClaudeLeibovici I understood sir. :) $\endgroup$ – danny Jul 22 '16 at 6:23
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A quadratic has $2$ equal roots when its determinant is equal to $0$. So we have

$$(k+1)^2-8k=0$$ $$k^2-6k+1=0$$

at which point you can solve using your preferred method.

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