The quadratic with a root of a root of a quadratic

Instead of directly using $x^2 - 79x + 1 = 0$ , we will simply use $x^2 - ax + 1 = 0$ as a generalization (assuming $a > 0$ of course), and then obtain a formula we can just plug in.

We present two solutions:

Solution 1: $r^2-ar+1 = 0$ $a = \frac{r^2+1}{r}$ $r-b\sqrt{r}+1 = 0$ $b = \frac{r+1}{\sqrt{r}}$ $b^2 = \frac{r^2+2r+1}{r}$ $b^2 = a+2$ $b = \boxed{\sqrt{a+2}}$

Solution 2: $(x^2-bx+1)(x^2+bx+1) = x^4 - (b^2-2)x^2 + 1 = 0$ $b^2 - 2 = a$ $b = \boxed{\sqrt{a+2}}$

Plugging in our value of $a$ , we get that $b = \sqrt{79+2} = \boxed{9}$ .

In both equations the product of the roots is 1. So the first equation has roots r and $\frac{1}{r}$ while the second has roots $\sqrt{r}$ and $\frac{1}{\sqrt{r}}$ .

Considering sums of roots we have $r + \frac{1}{r} = a$ and $\sqrt{r}$ + $\frac{1}{\sqrt{r}}$ = b

Squaring the second of these equations yields $r + \frac{1}{r} + 2 = b^2$ and therefore $a + 2 = b^2$

Substituting a=79 and taking the positive square root gives $b = 9$

Given that $r$ is a root of $x^2 - ax+1 = 0$ , therefore, $r^2 -ar + 1 = 0$ and $\implies a = \dfrac {r^2+1}r$ . Similarly, given that $\sqrt r$ is a root of $x^2 - bx+1 = 0$ , then

$\begin{aligned} b & = \frac {r+1}{\sqrt r} \\ & = \sqrt{\frac {(r+1)^2}r} \\ & = \sqrt{\frac {r^2+2r+1}r} & \small \color{#3D99F6} \text{Note that }a = \frac {r^2+1}r \\ & = \sqrt{a+2} & \small \color{#3D99F6} \text{and }a = 79 \\ & = \sqrt{79+2} \\ & = \boxed{9} \end{aligned}$

$r^2-79r+1=0$ $(r^2+2r+1)-81r=0$ $(r+1)^2=81r$ Since $r>0$ , taking square-root on both sides, we get $r+1=9\sqrt{r}$ $\sqrt{r}^2-9\sqrt{r}+1=0$ So $b=9$

Product of the roots is=1 fir both equations, the sum of the roots give relation b^2-2=79, therefore,

Answer is, b=9

We will use the fact that for a monic quadratic equation the sum of the roots is given by the negative of the coefficient of $x$ and the product of the roots is given by the constant term. Let R and r be the bigger and smaller root of the first equation, then

$R+r=79 \\ Rr=1$

The second equation with smaller root $\rho$ gives

$\sqrt{R} + \rho = b \\ \sqrt{R} \rho = 1$

Using these pairs of equations to eliminate the smaller roots gives

$79=R+\dfrac{1}{R} \dots(1) \\ b=\sqrt{R}+ \dfrac{1}{\sqrt{R}}$

Squaring this last equation gives

$b^2=R+ \dfrac{1}{R}+2$

Then using equation (1) gives

$b^2=79+2=81$

And so b (we are told t is positive) is $\boxed{9}$

Gentle reminder that Desmos exists

Got the correct answer, but did it the brute force way.

I only wanted to comment that you may want to be more precise with your definitions. Specifically, $x^2-ax+1=0$ has real solutions only when $\lvert a \rvert \geq 2$ ; the argument is the same when you replace $a$ with $b$ .

Aside from that, nice problem!