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

Question

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?

Answer 1

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 $p=1$.

Answer 2

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?

Answer 3

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 $p = -2$