Thread: Find the Roots

1. Find the Roots

As the title states, find the roots of

x^2 + (k^2 +1 all over k)x + 1 = 0

I would perfer a solution using the quadratic formula.

2. Originally Posted by Seronix
As the title states, find the roots of

x^2 + (k^2 +1 all over k)x + 1 = 0

I would perfer a solution using the quadratic formula.

Well first what are a,b, and c?

$\displaystyle a =1, b=\frac{k^2+1}{k}, c=1$

Then using the quadratric formula we get

$\displaystyle x=\frac{-\left( \frac{k^2+1}{k}\right) \pm \sqrt{\left( \frac{k^2+1}{k}\right)^2-4 }}{2(1)}$

Lets focus on the radicand for a second

$\displaystyle \left( \frac{k^2+1}{k} \right)^2-4=\frac{k^4+2k^2+1}{k^2}-4=\frac{k^4+2k^2+1}{k^2}-\frac{4k^2}{k^2}=\frac{k^4-2k^2+1}{k^2}$

$\displaystyle = \frac{(k^2-1)^2}{k^2}$

Now back to the radical

$\displaystyle x=\frac{-\left( \frac{k^2+1}{k}\right) \pm \sqrt{\left( \frac{(k^2-1)^2}{k^2}\right)^2}}{2(1)}= \frac{-\left( \frac{k^2+1}{k}\right) \pm \left( \frac{(k^2-1)^2}{k}\right)}{2}$

Well that is the hard part.

Good luck from here.

3. Hello, Seronix!

$\displaystyle x^2 + \left(\frac{k^2 +1}{k}\right)x + 1 \:= \:0$

We have: .$\displaystyle x^2 + \left(k + \frac{1}{k}\right)x + 1 \;=\;0$

Factor: .$\displaystyle (x + k)\left(x + \frac{1}{k}\right) \:=\:0$

Therefore: .$\displaystyle \begin{array}{ccccccc}x + k & = & 0 & \Rightarrow & x &=&\text{-}k \\ \\ [-2mm] x + \frac{1}{k} &=&0 & \Rightarrow & x &=& \text{-}\frac{1}{k} \end{array}$

4. Originally Posted by Soroban
Hello, Seronix!

We have: .$\displaystyle x^2 + \left(k + \frac{1}{k}\right)x + 1 \;=\;0$

Factor: .$\displaystyle (x + k)\left(x + \frac{1}{k}\right) \:=\:0$

Therefore: .$\displaystyle \begin{array}{ccccccc}x + k & = & 0 & \Rightarrow & x &=&\text{-}k \\ \\ [-2mm] x + \frac{1}{k} &=&0 & \Rightarrow & x &=& \text{-}\frac{1}{k} \end{array}$

Um yes, I solved the problem the same way. My problem is my teacher said to use the quadratic formula. It would seem ridiculous to do so since it requires so much work, but I guess i'm a bit curious on how to solve it using the quadratic formula as well. But Thank you nonetheless to both of you =)

5. Hello, Seronix!

Find the roots of: .$\displaystyle x^2 + \frac{k^2 +1}{k}x + 1 \:= \:0$
Multiply through by $\displaystyle {\color{blue}k}\!:\;\;kx^2 + (k^2+1)x + k \:=\:0$

Quadratic Formula: . $\displaystyle x \;=\;\frac{-(k^2+1) \pm \sqrt{(k^2+1)^2 - 4k^2}}{2k}$

. . $\displaystyle x \;=\;\frac{-(k^2+1) \pm\sqrt{k^4 + 2k^2 + 1 - 4k^2}}{2k} \;=\;\frac{-(k^2+1) \pm\sqrt{k^4-2k^2+1}}{2k}$

. . $\displaystyle x \;= \;\frac{-(k^2+1) \pm\sqrt{(k^2-1)^2}}{2k} \;=\;\frac{-(k^2+1) \pm (k^2-1)}{2k}$

And we have two roots:

. . $\displaystyle x \;=\;\frac{-(k^2 + 1) + (k^2-1)}{2k} \;=\;\frac{-k^2-1+k^2-1}{2k} \;=\;\frac{-2}{2k} \;=\;\boxed{-\frac{1}{k}}$

. . $\displaystyle x \;=\;\frac{-(k^2+1) - (k^2-1)}{2k} \;=\;\frac{-k^2-1 -k^2 + 1}{2k} \;=\;\frac{-2k^2}{2k} \;=\;\boxed{-k}$