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
so the roots are
$\displaystyle x = \beta$
$\displaystyle x = \beta + 2$
so your factorised expression is $\displaystyle (x- \beta)(x - (\beta +2))$
so $\displaystyle (x- \beta)(x - (\beta +2)) = x^2 + ax + (a + 2) $
expand giving $\displaystyle x^2 + +x( -2 \beta - 2) + \beta (\beta +2) = x^2 + ax + (a + 2) $
compare terms of same order
giving $\displaystyle -2 \beta - 2 = a$
and$\displaystyle \beta (\beta +2) = a + 2$
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.
A slightly different way.
We know that
$\displaystyle r_1 = \frac{-a}{2} + \sqrt{a^2 - 4(a + 2)}$
and
$\displaystyle r_2 = \frac{-a}{2} - \sqrt{a^2 - 4(a + 2)}$
by the quadratic formula, where r1 and r2 are the solutions to the quadratic.
Thus
$\displaystyle r_1 - r_2 = \sqrt{a^2 - 4(a + 2)} = 2$
Solve for a.
-Dan