Quadratic Equations: The Problem with Roots (1 Viewer)

[FDownes]

I gots me a question that I just can't seem to solve. I was hoping someone here could help me out by explaining it to me. So here it is;

Find the value of n for which the equation (n + 2)x^2 + 3x - 5 = 0 has one root triple the other.

EDIT: And another one;

Find the values of p for which x^2 - x + 3p - 2 > 0 for all x.

 

[Slidey]

x^2 + (3/(n+2))x - 5/(n+2) = 0

a+3a = -3/(n+2)
a*3a = -5/(n+2)

4a = -3/(n+2)
3a^2 = -5/(n+2)

Simultaneous equation, so isolate 1/(n+2):
4a/-3 = 1/(n+2)
3a^2/-5 = 1/(n+2)
So,
4a/-3 = 3a^2/-5
20a=9a^2
9a^2-20a=0
a(9a-20)=0
a=0 or a= 20/9
Sub into one of the initial equations (I'd suggest the 4a = ... one) to get:
Inconsistency, and:
-80/27=1/(n+2)
n=-197/80

Sub this back in: -(27/80)x^2 + 3x - 5 = 0 and solve to get roots of x= 2.22 and 6.66

Lovely.

 

[Slidey]

Find the values of p for which x^2 - x + 3p - 2 > 0 for all x.

I won't do it, but I'll give you hints:

You want to make the equation have NO roots. How do you figure out how many roots there are?

Discriminant.

Take the discriminant and find the value of p that makes it less than zero.

 

[FDownes]

Yes! Thank you very very much! I've been puzzling over that first question for ages!

EDIT: Got the second one too. Pretty easy once you think about it, I was just being lazy.

 

[FDownes]

Damn, stuck again. Here's the problem;

Solve 2^(2x +1) - 5 x 2^x + 2 = 0

I can see it's an equation reducible to a quadratic, the problem is the form it's in...

 

[watatank]

2^2x+1 - 5.2^x + 2 = 2.(2^2x) - 5.2^x + 2 = 2.(2^x)² - 5.2^x + 2

let 2^x be "a"

then you have a simple quadratic.