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Finding the quadratic equation that suits the solutions

  1. Nov 15, 2015 #1
    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. jcsd
  3. Nov 15, 2015 #2

    pasmith

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    Homework Helper

    Observation: [itex]x_1[/itex] and [itex]x_2[/itex] are the solutions of [itex]x^2 + bx + c = 0[/itex] if and only if [itex]x_1 + x_2 = -b[/itex] and [itex]x_1x_2 = c[/itex], since [tex]
    (x - x_1)(x - x_2) = x^2 - (x_1 + x_2)x + x_1x_2.[/tex] You didn't need to solve [itex]x^2 + x + 1 = 0[/itex]; the quadratic you are looking for is [tex]y^2 - (y_1 + y_2)y + y_1y_2 = 0,[/tex] and both [itex]y_1 + y_2[/itex] and [itex]y_1y_2[/itex] can be expressed in terms of [itex]a[/itex], [itex]x_1 + x_2 = -1[/itex] and [itex]x_1x_2 = 1[/itex] by adding and multiplying equations (2) and (3) respectively.
     
  4. Nov 15, 2015 #3
    That is really helpful, now i have solved it rightly. Its y^2 - (a+1)y + a^2 - a + 1 = 0
    Thanks
     
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