Algebra

**sparky69er** · December 29th 2007, 05:35 AM

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

**bobak** · December 29th 2007, 06:07 AM

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.

**topsquark** · December 29th 2007, 06:52 AM

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

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