# Thread: [SOLVED] Miscellaneous 6 OCR

1. ## [SOLVED] Miscellaneous 6 OCR

Hi everyone, I am new in this forum. Anyway I have been struggling with an inequality question in my homework, unfortunately due in tomarrow.

Find the value of k for the following equation which has 2 real roots

Kx^2 + kx + 2 = 0

Now I have found that: b^2 -4ac
k^2 - 4*k*2> 0
K^2 -8k > 0

I do not understand what to do after this stage because it is not possible to expand the brackets or square root the 8

2. Originally Posted by james singh jagjit
Hi everyone, I am new in this forum. Anyway I have been struggling with an inequality question in my homework, unfortunately due in tomarrow.

Find the value of k for the following equation which has 2 real roots

Kx^2 + kx + 2 = 0

Now I have found that: b^2 -4ac
k^2 - 4*k*2> 0
K^2 -8k > 0

I do not understand what to do after this stage because it is not possible to expand the brackets or square root the 8
$\displaystyle b^2 - 4ac > 0$

$\displaystyle k^2 - 4(k)(2) > 0$

$\displaystyle k( k - 8 ) > 0$

Thus

---0_________8---

$\displaystyle k < 0 \ and \ 8 < k$

---
Thanks for the spot Jhevon!

3. Hello, James!

Welcome aboard!

Find the value of $\displaystyle k$ for which the following equation
has two real roots: .$\displaystyle kx^2 + kx + 2 \:= \:0$

Now I have found that: .$\displaystyle b^2 -4ac \:> \;0$

So we have: .$\displaystyle k^2 -8k \:> \:0$
Factor: .$\displaystyle k(k-8) \:>\:0$

It says: The product of two numbers is positive.
When does this happen?
. . It happens in two ways . . .

[1] Both factors are positive: .$\displaystyle \begin{Bmatrix}k & > & 0 & \Rightarrow & k & > & 0 \\ k - 8 & > & 0 & \Rightarrow & k & > & 8 \end{Bmatrix}\quad\Rightarrow\quad k \:> \:8$

[2] Both factors are negative: .$\displaystyle \begin{Bmatrix}k & < & 0 & \Rightarrow & k & < & 0 \\ k-8 & <& 0 & \Rightarrow & k &<& 8 \end{Bmatrix}\quad\Rightarrow\quad k \:<\:0$

Therefore, the equation has two real roots for: .$\displaystyle k\,\in\,(-\infty,0) \cup (8,\infty)$