1. ## Algebra

Given that the roots of the equation x^2 + ax + (a + 2) = 0
differ by 2, find the possible values of the constant a.

ans= -2 and 6

Can anyone show me how to work out this question please?

cheers

2. so the roots are
$x = \beta$
$x = \beta + 2$

so your factorised expression is $(x- \beta)(x - (\beta +2))$

so $(x- \beta)(x - (\beta +2)) = x^2 + ax + (a + 2)$

expand giving $x^2 + +x( -2 \beta - 2) + \beta (\beta +2) = x^2 + ax + (a + 2)$

compare terms of same order

giving $-2 \beta - 2 = a$
and $\beta (\beta +2) = a + 2$

so all you need to do is solve this system to find the two values of a.

Note an alternative method is to use the quad formula.

3. Originally Posted by sparky69er
Given that the roots of the equation x^2 + ax + (a + 2) = 0
differ by 2, find the possible values of the constant a.

ans= -2 and 6

Can anyone show me how to work out this question please?

cheers
A slightly different way.

We know that
$r_1 = \frac{-a}{2} + \sqrt{a^2 - 4(a + 2)}$
and
$r_2 = \frac{-a}{2} - \sqrt{a^2 - 4(a + 2)}$
by the quadratic formula, where r1 and r2 are the solutions to the quadratic.

Thus
$r_1 - r_2 = \sqrt{a^2 - 4(a + 2)} = 2$

Solve for a.

-Dan