# Roots of a Quadraitc Equation

• Jul 3rd 2013, 10:01 AM
Seaniboy
Hi,
I would appreciate any help with the following question.

Given that the roots of the equation x² + ax + (a + 2) = 0 differ by two, find the possible values of constant a.
• Jul 3rd 2013, 10:29 AM
ebaines
Re: Roots of a Quadraitc Equation
Use the quadratic equation to get an expression for the two roots of the equation, and set the difference between them equal to two:

$\displaystyle x_1 = \frac {-a + \sqrt{a^2-4(a+2)}}2$

$\displaystyle x_2 = \frac {-a - \sqrt{a^2-4(a+2)}}2$

subtract, and set equal to 2:

$\displaystyle x_1 - x_2 = \sqrt{a^2-4(a+2)} = 2$

Now solve for a - square both sides, rearrange, and use the quadratic formula.
• Jul 3rd 2013, 03:19 PM
HallsofIvy
Re: Roots of a Quadraitc Equation
Another way: $\displaystyle (x- x_0)(x- x_1)= x^2- (x_0+ x_1)x+ x_0x_1= 0$ so if $\displaystyle x_0$ and $\displaystyle x_1$ are roots of $\displaystyle x^2+ ax+ (a+ 2)= 0$ then we must have $\displaystyle x_0+ x_1= a$ and $\displaystyle x_0x_1= a+ 2$. Knowing also that "the roots of the equation x² + ax + (a + 2) = 0 differ by two", so that $\displaystyle x_1= x_0+ 2$, we have three equations to solve for the three values, $\displaystyle x_0$, $\displaystyle x_1$, a.
• Jul 4th 2013, 12:35 PM
Seaniboy
Re: Roots of a Quadraitc Equation
Hi,
Thanks very much for your help.
Much appreciated.
• Jul 4th 2013, 01:09 PM
Seaniboy
Re: Roots of a Quadraitc Equation
Thanks a lot for your help.