# Finding the quadratic equation that suits the solutions

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1. Nov 15, 2015

### diredragon

1. The problem statement, all variables and given/known data
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?

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

3. 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.

2. Nov 15, 2015

### 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 $$(x - x_1)(x - x_2) = x^2 - (x_1 + x_2)x + x_1x_2.$$ 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.

3. Nov 15, 2015

### diredragon

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