SOLUTION: If the roots are ax^2+bx+c=0, differ by 1, show that they are (a-b)/2a and -(a+b)/2a, and prove that b^2=a(a+4c)

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: If the roots are ax^2+bx+c=0, differ by 1, show that they are (a-b)/2a and -(a+b)/2a, and prove that b^2=a(a+4c)      Log On


   

Question 1132708: If the roots are ax^2+bx+c=0, differ by 1, show that they are (a-b)/2a and -(a+b)/2a, and prove that b^2=a(a+4c)
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
If the roots are ax^2+bx+c=0, differ by 1,
Let the roots be r and r+1

ar%5E2%2Bbr%2Bc=0 and a%28r%2B1%29%5E2%2Bb%28r%2B1%29%2Bc=0

So ar%5E2%2Bbr%2Bc=a%28r%2B1%29%5E2%2Bb%28r%2B1%29%2Bc

ar%5E2%2Bbr%2Bc=a%28r%5E2%2B2r%2B1%29%5E2%2Bbr%2Bb%2Bc
 
ar%5E2%2Bbr%2Bc=ar%5E2%2B2ar%2Ba%2Bbr%2Bb%2Bc

0=2ar%2Ba%2Bb

2ar%2Ba%2Bb=0

2ar=+-a-b

r+=+%28-a-b%29%2F%282a%29   <-- that's one root

The other root is r+1, so we add 1 to both sides:

r%2B1+=+%28-a-b%29%2F%282a%29%2B1

r%2B1+=+%28-a-b%29%2F%282a%29%2B%282a%29%2F%282a%29

r%2B1+=+%28-a-b%2B2a%29%2F%282a%29

r%2B1+=+%28a-b%29%2F%282a%29  <-- that's the other root

and prove that b^2=a(a+4c)
Substitute either root in the original.  I'll use

%28a-b%29%2F%282a%29, since it has only one - sign. 

a%28%28a-b%29%2F%282a%29%29%5E2%2Bb%28%28a-b%29%2F%282a%29%29%2Bc=0

a%28%28a%5E2-2ab%2Bb%5E2%29%2F%284a%5E2%29%29%2B%28%28ab-b%5E2%29%2F%282a%29%29%2Bc=0 



%28%28a%5E2-2ab%2Bb%5E2%29%2F%284a%29%29%2B%28%28ab-b%5E2%29%2F%282a%29%29%2Bc=0

Multiply through by 4a

%28a%5E2-2ab%2Bb%5E2%29%2B%282ab-2b%5E2%29%2B4ac=0

a%5E2-2ab%2Bb%5E2%2B2ab-2b%5E2%2B4ac=0

a%5E2-b%5E2%2B4ac=0

a%5E2%2B4ac=b%5E2

a%28a%2B4c%29=b%5E2

b%5E2=a%28a%2B4c%29

Edwin