Sum and Product of the Roots (Quadratic Equations)

[zebra1707]

Homework Statement

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

Homework Equations

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

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

 

[Mark44]

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

Find p if twice the sum of the roots and product of roots

Part of the sentence is missing.

 

[willem2]

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]

Hi there

I have edited the original question - my apologies there.

Cheers

 

[zebra1707]

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. Homework 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]

1) No the sum is $-b/a=-(-p/1)=p$ 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]

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]

Yes you were very close. You had the right approach.
You just forgot about the squaring p in the $\Delta=(-p)^2-4.1.9$

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 $\Delta<0$ 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 $p^2\neq 36$ thus, $p\neq \pm 6$ (Note: do not forget about the plus/minus)

 

[zebra1707]

Many thanks

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

Cheers