# Thread: Equations with Common Roots

1. ## Equations with Common Roots

1. Show that, if the equations x^2 + 2px + q = 0 and x^2+2Px+Q=0 have a common root, then (q-Q)^2+4(P-p)(Pq-pQ)=0.

2. Find the conditions that the equations x^2+2x+a=0 and x^2+bx+3 should have a common root.
Hi guys
I have no clue how to do these problems.
Could someone please help???

2. Originally Posted by xwrathbringerx
1. Show that, if the equations x^2 + 2px + q = 0 and x^2+2Px+Q=0 have a common root, then (q-Q)^2+4(P-p)(Pq-pQ)=0.

2. Find the conditions that the equations x^2+2x+a=0 and x^2+bx+3 should have a common root.
If there is a value of x for which $x^2 + 2px + q = 0$ and $x^2+2Px+Q=0$ then $2px+q = 2Px+Q$. Solve that equation for x, then plug that value of x into one of the quadratic equations and simplify the result. That will give you the solution to 1. For 2., all you have to do it is to put the values p=2, q=a, P=b, Q=3 into the result from 1.

3. Hmmm for 1. i can't seem to get the equation as shown in the equation. The equation is like (Q-q)^2/(2p-2P)^2 + 2P(Q - q)/(2p-2P) + Q = 0. When I expand or attempt to contract it, I dont seem to get anywhere closer to the answer.

For 2. How did u find out the values?

4. ## here is the method

let a be common root therfor
a^2+2pa+q=0
a^2+2Pa+Q=0
we will apply the method of cross multiplication so

2p q 1 2p
2P Q 1 2P

(a^2)/(2pQ-2Pq) = a/(q-Q) = 1/(2P-2p)

from above
a=(2pQ-2Pq) /(q-Q) = (q-Q)/(2P-2p)

now cross multiplying we get
4(P-p)(pQ-Pq)=(q-Q)^2
or
(q-Q)^2 +4(P-p)(Pq-pQ)=0

NOTE : this is also the method to find the condition of common root so your second question will be solved in the same way.

5. Originally Posted by Opalg
If there is a value of x for which $x^2 + 2px + q = 0$ and $x^2+2Px+Q=0$ then $2px+q = 2Px+Q$. Solve that equation for x, then plug that value of x into one of the quadratic equations and simplify the result. That will give you the solution to 1. For 2., all you have to do it is to put the values p=2, q=a, P=b, Q=3 into the result from 1.
Hmmmmm for the values of P and p ... so it's not 2p = a or 2P = b, just p = a and P = b????

6. Originally Posted by xwrathbringerx
Hmmmmm for the values of P and p ... so it's not 2p = a or 2P = b, just p = a and P = b????
Yes, I was wrong. I forgot about the 2s. So it should be 2p=2 and 2P=b. (But you're wrong too! It's 2p=2 and q=a, not 2p=a.)