[SOLVED] Miscellaneous 6 OCR

• Nov 26th 2007, 07:55 AM
james singh jagjit
[SOLVED] Miscellaneous 6 OCR
Hi everyone, I am new in this forum:cool:. Anyway I have been struggling with an inequality question in my homework, unfortunately due in tomarrow:mad:.

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
• Nov 26th 2007, 08:53 AM
janvdl
Quote:

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

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

$b^2 - 4ac > 0$

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

$k( k - 8 ) > 0$

Thus

---0_________8---

$k < 0 \ and \ 8 < k$

---
Thanks for the spot Jhevon! (Handshake)
• Nov 26th 2007, 09:35 AM
Soroban
Hello, James!

Welcome aboard!

Quote:

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

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

So we have: . $k^2 -8k \:> \:0$

Factor: . $k(k-8) \:>\:0$

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

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

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

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