Thread: Find a and m in this equation: x^2+2(m-a)x+3am-2=0

1. Find a and m in this equation: x^2+2(m-a)x+3am-2=0

There is the following equation:
x^2+2(m-a)x+3am-2=0 | a,m ∈ R
a) Find a such that the euation has real roots, ∀ m ∈ R.
b) Find m such that the euation has real roots, ∀ a ∈ R.
I put the condition: DELTA >= 0
I found DELTA but since this step I have problems..
Sorry for my english!
The answer should be: |a| <= sqrt(8/21) ; |m| <= sqrt(8/21)
My try:

2. idea

Can anyone give me a little help or an idea?

3. Originally Posted by Vali
There is the following equation:
x^2+2(m-a)x+3am-2=0 | a,m ∈ R
a) Find a such that the equation has real roots, ∀ m ∈ R.
b) Find m such that the equation has real roots, ∀ a ∈ R.
I put the condition: DELTA >= 0

For what does "DELTA" stand? How does it relate to the posted question?

Originally Posted by Vali
I found DELTA but since this step I have problems.
What were your steps? What did you obtain? How did this relate to "a" and "m"?

Originally Posted by Vali
The answer should be: |a| <= sqrt(8/21) ; |m| <= sqrt(8/21)
My try:

The image in the attachment appears to show the following:

$\Delta\, =\, h\, (a^2\, -\, 5am\, +\, m^2\, +\, 2)$

$a^2\, -\, 5am\, +\, m^2\, +\, 2\, \geq\, 0$

$\Delta '\, \geq\, 0$

$\Delta '\, =\, 25m^2\, -\, 3m^2\, -\, 8\, =\, )$

$\Delta '\, =\, 21m^2\, -\, 8$

$21m^2\, -\, 8\, \geq\, 0$

$21m^2\, \geq\, 8\, =\, )$

$\big|m\big|\, \geq\, \sqrt{\dfrac{8}{21}\,}$

Please reply with confirmation or corrections, along with a clear explanation of what you're doing. Thank you!

Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•