A maths - quadratic equation

• bonkysleuth

  25 Apr 08, 17:11

  given the equation (p-1)x^2 - 3x + 10 + p =0 where p>0, find the values of p for which one root is twice the other..

  let roots be alfa and 2 alfa ( represented by the bold a)

  and so i got 3a = 3/(p-1)    - - - (1)

  2a^2 = 10+p / p-1               - - - (2)

   

  From (1)

  a : 3/p-1 X 1/3

  After substituting the a from (1) into (2) and doing the cross-multiplication and all, i got

   

  p^3 8p^2 - 21p + 12 = 0

   

  I can't get p from this equation. can someone help? the answer's 1.18

• ^tamago^

  25 Apr 08, 17:38

  (p-1)x²-3x+(10+p)=(x-α)(x-2α)=x²-3αx+2α²=0 x²-[3/(p-1)]x+[(p+10)/(p-1)]=x²-3αx+2α² By comparing, 1=α(p-1) 1=α²(p-1)² ––– (1) 2α²(p-1)=p+10 p+10=2α²(p-1) ––– (2) (1) ÷ (2): p-1 = 2/(p+10) (p-1)(p+10) = 2 p² + 9p -12 = 0 p = 1.1789083 ≈ 1.18 (3 s.f.)