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For what real values of $k$ does $x^2+(k+1)x+k^2$ have one root double the other?

For a start, I found the range of $k$ which endows this equation with real roots: $$-\frac13\le k\le1$$

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  • $\begingroup$ But this is not a cubic equation... $\endgroup$ – boaz Oct 17 '16 at 16:30
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    $\begingroup$ If the roots are $r,2r$ then your polynomial is $p(x)=(x-r)(x-2r)$ so $k^2=2r^2$ and $k+1=-3r$. $\endgroup$ – lulu Oct 17 '16 at 16:35
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Suppose the equation has one root double the other. Then it can be written as $$(x-a)(x-2a)=x^2-3ax+2a^2$$ Comparing coefficients between this expression and $x^2+(k+1)x+k^2$ we have $$k+1=-3a\qquad2a^2=k^2$$ From the first equation we have $k=-3a-1$; substituting this into the second equation yields $$2a^2=(-3a-1)^2=9a^2+6a+1$$ $$7a^2+6a+1=0$$ $$a=\frac{-6\pm\sqrt{6^2-4\cdot7\cdot1}}{2\cdot7}=\frac{-3\pm\sqrt2}7$$ From here we can recover the possible values of $k$: $$k=-3\left(\frac{-3\pm\sqrt2}7\right)-1=\frac{2\pm3\sqrt2}7$$

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  • $\begingroup$ how do you compare coefficients? $\endgroup$ – Anne Oct 17 '16 at 16:59
  • $\begingroup$ @Anne If one polynomial is in the form of another, their coefficients in the constant term, linear term, quadratic term, etc. must agree. Just compare term by term. $\endgroup$ – Parcly Taxel Oct 17 '16 at 17:01
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Let $a$ and $b$ be the roots. Since $$ (x-a)(x-b)=x^2+(k+1)x+k^2\\ x^2-(a+b)x+ab=x^2+(k+1)x+k^2 $$ it follows (by comparing corresponding coefficients) that $$ \begin{cases} a+b=-(k+1) & (1)\\ ab=k^2 & (2) \end{cases} $$ If $a=2b$, then $$ \begin{cases} 3b=-(k+1) & (3)\\ 2b^2=k^2 & (4) \end{cases} $$ Flugging (3) inside (4) gives $$ \sqrt{2}\left|-\frac{k+1}{3}\right|=|k| $$ so $k=\frac{3}{7}\sqrt{2}\pm\frac{2}{7}$.

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  • $\begingroup$ How do you get the first sytem of equations? $\endgroup$ – Anne Oct 17 '16 at 16:53
  • $\begingroup$ It called "Vieta's formulas" (here i used the quadratic case). You can get it directly by comparing the corresponding coefficients. I reedit my answer for more clarity. $\endgroup$ – boaz Oct 17 '16 at 16:57
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Let $x_1$ and $x_2$ be the two roots and $x_1 = 2x_2$. Then we have by Vieta's formulae: $-(k+1_ = x_1 + x_2 = 3x_2 \implies x_2 = -\frac{1+k}{3}$. On the other hand: $k^2 = x_1x_2 = 2x_2^2 \implies x_2 = \pm \frac{k}{\sqrt{2}}$

Now equate the two equation and you will be able to solve for $k$ easily..

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    $\begingroup$ Should be $-(k+1)$, not $-k+1$. $\endgroup$ – Thomas Andrews Oct 17 '16 at 16:39
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Let the given polynomial be, $f(x)$ and the roots be, $\alpha, 2\alpha.$

Then, $f(x)= x^2-(\alpha + 2\alpha)+\alpha \cdot 2\alpha =x^2-3\alpha+2\alpha ^2$ (How?)

$\therefore k^2=2\alpha ^2 $

$\implies k=\pm \sqrt{2}\alpha ...(1)$

Again, $k+1=-3\alpha...(2)$

Dividing (2) by (1), we get

$$\frac{k+1}{k}=\frac{-3}{\pm\sqrt{2}}$$

The solution of the equation is left to you. I hope it is easy to continue from here.

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