Finding the quadratic equation that suits the solutions

[diredragon]

Homework Statement

if x1 and x2 are solutions to 1) x² + x + 1 = 0, then 2) y1 = ax1 + x2 and 3) y2 = x1 + ax2 are solutions to which quadratic equation?

Homework Equations

ax² + bx + c = 0
x1∕2 = (−b ± √(b² - 4ac))/2a

The Attempt at a Solution

Well, firstly i solved for x1 and x2 getting:
x1 = (−1 + √(-3))/2
x2 = (−1 - √(-3))/2
I now have in mind to replace the x1 and x2 in equations 2) and 3), and after i simplify best i can to find the equation by means of following formula:
(y - y1)(y - y2)
Will this yield the solution, and is there a simpler way of doing this? The simplification alone is a hard task letalone the multiplication that comes afterwards.
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[pasmith]

Observation: x_1 and x_2 are the solutions of x^2 + bx + c = 0 if and only if x_1 + x_2 = -b and x_1x_2 = c, since <br /> (x - x_1)(x - x_2) = x^2 - (x_1 + x_2)x + x_1x_2. You didn't need to solve x^2 + x + 1 = 0; the quadratic you are looking for is y^2 - (y_1 + y_2)y + y_1y_2 = 0, and both y_1 + y_2 and y_1y_2 can be expressed in terms of a, x_1 + x_2 = -1 and x_1x_2 = 1 by adding and multiplying equations (2) and (3) respectively.

 

[diredragon]

That is really helpful, now i have solved it rightly. Its y^2 - (a+1)y + a^2 - a + 1 = 0
Thanks