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.
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.
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.