Solve the quadratic equation (a+b-2c)x² + (2a-b-c)x + (c+a-2b) =0 is
-B = -2a+b+c
B^2 = 4a^2-4ab-4ac+b^2+2bc+c^2
-4AC = -4a^2+4ab+4ac+8b^2-20bc+8c^2
so
B^2-4AC=9b^2-18bc+9c^2 = 9(b^2-2bc+c^2)
=9(b-c)^2 whew!
sqrt(B^2-4AC) = 3(b-c) = 3b - 3c
so
-B+/-sqrt(B^2-4AC) = -2a+b+c +/- (3b-3c)
= -2a+4b-2c or -2a-2b+4c
divide those by 2A
To solve the quadratic equation (a+b-2c)x² + (2a-b-c)x + (c+a-2b) = 0, we can use the quadratic formula. The quadratic formula states that for an equation of the form ax² + bx + c = 0, the solutions for x can be found using the formula:
x = (-b ± √(b² - 4ac)) / (2a)
In our case, we have the equation (a+b-2c)x² + (2a-b-c)x + (c+a-2b) = 0, where a, b, and c are coefficients.
Comparing it to the general form ax² + bx + c = 0, we can determine the values of a, b, and c as follows:
a = a + b - 2c
b = 2a - b - c
c = c + a - 2b
Now we can substitute these values into the quadratic formula to find the solutions for x.