x^2 + a(3a-5)x = 2(x+4a)
x^2 -2x-8a + +3a^2x - 5ax = 0
x^2 + (3a^2-5a-2)x + 3a^2-8a = 0
since the sum of the roots is -b/a, we have 3a^2-5a-2 = 0.
(3a+1)(a-2) = 0
You can probably take it from here.
a)If one root of the equation
x^2 + a(3a-5)x = 2(x+4a) is negative of the other. Find the values of a .
b) If a>0, use the result of (a) to solve the equation.
1 answer
1 answer