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If one root of the equation $x^2+px+1=0$ be the square root of the other then find the value of $p$.

My Attempt:

Let $\alpha $ and $\beta $ be the two roots of the given equation. Then,

$$\alpha =\dfrac {-p + \sqrt {p^2-4}}{2}$$ $$\beta =\dfrac {-p-\sqrt {p^2-4}}{2}$$.

According to Question: $$\alpha =\sqrt {\beta}$$ $$2p^2-2p\sqrt {p^2-4} - 4=-2p-2\sqrt {p^2-4}$$

How do I proceed further?

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  • $\begingroup$ Why do you assume alpha is the square root of beta and nor the other way around? $\endgroup$ – fleablood Apr 26 '17 at 5:54
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The roots are $a$ and $a^2$ so $$a^3=1$$ and $$a+a^2=-p$$ so either $a=1$ and $p=-2$ or $$a^2+a+1=0$$ in which case $p=1$.

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  • $\begingroup$ How is.$a^3=1$? $\endgroup$ – pi-π Apr 26 '17 at 6:26
  • $\begingroup$ If the roots are $a$ and $a^2$ then the factors are $x-a$ and $x-a^2$. Multiply them together. $\endgroup$ – Leonard Blackburn Dec 6 '17 at 15:44
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So the roots are $r$ and $r^2$. Then the equation is $(x-r)(x-r^2)=0$, that is $x^2-(r+r^2)x+r^3=0$. So you need $p=-(r+r^2)$ and $1=r^3$. So what is $r$, and what then must $p$ be?

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  • $\begingroup$ why are $r$ and $r^2$ roots?..and please explain the other steps as well. $\endgroup$ – pi-π Apr 26 '17 at 5:49
  • $\begingroup$ @Maxwell if the first root is the square root of the second, then the second is the square of the first. $\endgroup$ – Lord Shark the Unknown Apr 26 '17 at 5:50
  • $\begingroup$ And how do you get $p=-(r+r^2)$ and $1=r^3$? $\endgroup$ – pi-π Apr 26 '17 at 5:58
  • $\begingroup$ If it makes you more comfortable (but it shouldn't-- this should both be equally clear) you could let one root be $t$ and the other $\sqrt t $. Then you must solve $\t\sqrt t=1$ and $p=-(t+\sqrt t) $. But that is the *exact* same thing with $t=r^2$. $\endgroup$ – fleablood Apr 26 '17 at 6:01
  • $\begingroup$ @fleablood, solving this way gives $p=-2$ but the answer in book is $1$. $\endgroup$ – pi-π Apr 26 '17 at 6:04
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Vieta's formulas:

a = 1, b = p, c = 1 for your equation ($x_1 = \sqrt{x_2}$)

$x_1x_2 = \frac{c}{a} = 1 = x\sqrt{x}$, thus $x = 1$.

$x_1 + x_2 = -\frac{b}{a} = -p$

$1 + 1 = -p$, thus $p = -2$

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Assuming the roots are $\alpha$ and $\alpha^2$, then, \begin{align} &(x-\alpha)(x-\alpha^2)=0 \\ \implies&x^2+x(-\alpha-\alpha^2)+\alpha^3=0 \end{align}

Comparing this with $x^2+px+1 = 0$ we get: \begin{align} &\alpha+\alpha^2 = -p \\ &\alpha^3=1 \end{align}

If we cube the first relation of the two above, we discover an interesting relation: \begin{align} &(\alpha+\alpha^2)^3=-p^3 \\ \implies&\alpha^3+\alpha^6+3\alpha^4+3\alpha^5=-p^3 \\ \implies&a^3+a^6+3\alpha^3(\alpha+\alpha^2) = -p^3 \\ \implies&1+1+3(-p)=-p^3 \\ \implies&p^3-3p+2=0 \\ \implies&(p-1)(p^2+p-2)=0 \\ \implies&(p-1)^2(p+2)=0 \end{align}

Clearly for p to satisfy this $p=1$ or $p=-2$

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