SOLUTION: im really stuck on this question the roots of the quadratic equation x^2 + 4x - a= 0 are b+1 and b-3. Find the values of a and b. thnks :)

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: im really stuck on this question the roots of the quadratic equation x^2 + 4x - a= 0 are b+1 and b-3. Find the values of a and b. thnks :)       Log On


   

Question 355091: im really stuck on this question
the roots of the quadratic equation x^2 + 4x - a= 0 are b+1 and b-3. Find the values of a and b.
thnks :)

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
One form of a quadratic equation is:
%28x+-+r%5B1%5D%29%28x+-+r%5B2%5D%29+=+0
In this form r%5B1%5D and r%5B2%5D are the two roots of the quadratic equation. This is the form which is often used when solving quadratic equations.

So you equation, with its roots of b+1 and b-3, can be written in the form:
%28x+-+%28b%2B1%29%29%28x+-+%28b-3%29%29+=+0
which simplifies to:
%28x+-+b+-+1%29%28x+-+b+%2B+3%29+=+0
Multiplying this out (multiplying each term of one factor by each term of the other) we get:
x%5E2+-bx+%2B+3x+-bx+%2B+b%5E2+-3b+-x+%2Bb+-+3+=+0
Combining like terms we get:
x%5E2+-2bx+%2B+2x+%2B+b%5E2+-2b+-+3+=+0
Then, for reasons that should be clear soon, I'll factor out the x from -2bx + 2x and also group the constant terms (the terms with no "x"):
x%5E2+%2B+%28-2b+%2B+2%29x+%2B+%28b%5E2+-2b+-+3%29+=+0
In order for this equation to be the same as:
x%5E2+%2B+4x+-+a=+0
then the coefficients of the x terms must be the same:
-2b + 2 = 4
and the constant terms must be the same:
b%5E2+-2b+-+3+=+-a
(The x%5E2 terms are already the same.) From
-2b + 2 = 4
we can solve for b:
-2b = 2
b = -1
Then we can use this and the equation b%5E2+-2b+-+3+=+a to find a:
%28-1%29%5E2+-2%28-1%29+-+3+=+-a
1+-2%28-1%29+-+3+=+-a
1+%2B2+-+3+=+-a
0+=+-a
0+=+a
So a = 0 and b = -1 and your equation was x%5E2+%2B+4x+=+0