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Sum and Product of the Roots (Quadratic Equations)

by zebra1707
Tags: equations, product, quadratic, roots
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zebra1707
#1
Oct12-09, 11:40 PM
P: 107
1. The problem statement, all variables and given/known data

For the quad equation x^2 - px + 9 = 0

1. Write down the sum of roots and product of roots
2. Find p IF twice the sum of the roots EQUALS the product
3. Find p IF the roots are unequal

2. Relevant equations

Sum = (a+b) = -b/a Product = (ab) c/a

3. The attempt at a solution

1. Using the formula Sum = -p Product = 9
2. -2p = 9 -9/2 = 2p/2 = 4 1/2
3. Totally lost

Can someone provide guidence. Cheers
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Mark44
#2
Oct13-09, 12:59 AM
Mentor
P: 21,216
Let r1 and r2 be the roots of the given quadratic.
1) Sum of roots = r1 + r2 = -p
Product of roots = r1 * r2 = 9

You have two equations in two unknowns. Can you solve for r1 and r2 in terms of p?

2) What's the question in this part? You have
Quote Quote by zebra1707
Find p if twice the sum of the roots and product of roots
Part of the sentence is missing.
willem2
#3
Oct13-09, 01:10 AM
P: 1,395
1 if the roots are a,b then the equation is x^2 - (a+b)x + ab = 0, so the coefficient of x is -(sum of the roots), and you should have p instead of -p.

2. I have no idea what is meant here.

3. Find p when the roots are equal first. Can you use comples numbers? if not there are more values of p where the quadratic doesn't have a solution

zebra1707
#4
Oct13-09, 01:11 AM
P: 107
Sum and Product of the Roots (Quadratic Equations)

Hi there

I have edited the original question - my apologies there.

Cheers
zebra1707
#5
Oct13-09, 01:16 AM
P: 107
I have amended my original post.

For the quad equation x^2 - px + 9 = 0

1. Write down the sum of roots and product of roots
2. Find p IF twice the sum of the roots EQUALS the product
3. Find p IF the roots are unequal

2. Relevant equations

Sum = (a+b) = -b/a Product = (ab) c/a

3. The attempt at a solution

1. Using the formula Sum = -p Product = 9
2. -2p = 9 -9/2 = 2p/2 = 4 1/2
3. Totally lost

Can someone provide guidence. Cheers
Mentallic
#6
Oct13-09, 05:05 AM
HW Helper
P: 3,502
1) No the sum is [itex]-b/a=-(-p/1)=p[/itex] and the product is right.

2) You're right except for taking the sum as -p rather than p.

3) If we need p when the roots are unequal, how about we find the value(s) of p when the roots are equal, then take all other values?
zebra1707
#7
Oct13-09, 08:58 PM
P: 107
I think that I have nutted out part 3, of this question

x^2 - px + 9 = 0

a = 1 b = -p and c = 9

Delta = b^2 - 4ac
= (-p)^2 - 4(1) (9)
= p - 36

So if plugged into the following:

Equal roots Delta = 0
p - 36 = 0
p = 36

For real roots Delta = >(Equal to) 0
p - 36 >(Equal to) 0

Unreal Delta < 0
p - 36 < 0
p < 36

For real and different Delta > 0
p - 36 > 0
p > 36

Guidence on this would be great
Mentallic
#8
Oct14-09, 03:36 AM
HW Helper
P: 3,502
Yes you were very close. You had the right approach.
You just forgot about the squaring p in the [itex]\Delta=(-p)^2-4.1.9[/itex]

However, there were no other restrictions on the problem. It just said find p when the roots are unequal. It never said anything about the roots being real/imaginary.
Basically, taking [itex]\Delta<0[/itex] is fine too. It just means for those values of p, the quadratic will be entirely above the x-axis.

So finally, for roots unequal, p is all reals except [itex]p^2\neq 36[/itex] thus, [itex]p\neq \pm 6[/itex] (Note: do not forget about the plus/minus)
zebra1707
#9
Oct14-09, 03:44 AM
P: 107
Many thanks

I understand - many thanks for taking the time to respond so thoughtfully.

Cheers


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