Question 355091
One form of a quadratic equation is:
{{{(x - r[1])(x - r[2]) = 0}}}
In this form {{{r[1]}}} and {{{r[2]}}} are the two roots of the quadratic equation. This is the form which is often used when solving quadratic equations.<br>
So you equation, with its roots of b+1 and b-3, can be written in the form:
{{{(x - (b+1))(x - (b-3)) = 0}}}
which simplifies to:
{{{(x - b - 1)(x - b + 3) = 0}}}
Multiplying this out (multiplying each term of one factor by each term of the other) we get:
{{{x^2 -bx + 3x -bx + b^2 -3b -x +b - 3 = 0}}}
Combining like terms we get:
{{{x^2 -2bx + 2x + b^2 -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^2 + (-2b + 2)x + (b^2 -2b - 3) = 0}}}
In order for this equation to be the same as:
{{{x^2 + 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^2 -2b - 3 = -a}}}
(The {{{x^2}}} 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^2 -2b - 3 = a}}} to find a:
{{{(-1)^2 -2(-1) - 3 = -a}}}
{{{1 -2(-1) - 3 = -a}}}
{{{1 +2 - 3 = -a}}}
{{{0 = -a}}}
{{{0 = a}}}
So a = 0 and b = -1 and your equation was {{{x^2 + 4x = 0}}}