given that x^2+2mx+m^2-2m+3=0 has 2 roots a and B (with a<B)
1. Find a+B and aB in terms of m
2. Find (a-b)^2 in terms of m
3. If B-a=2, find m
a+b = -2m
ab = (m^2-2m+3)
(a-b)^2 = (a+b)^2 - 4ab
now finish it off
oobleck how about number 3?
I recommend asking your parents
Hh, I don't have my parents with me.
They are with you.
oobleck is it 8m-12
To find a and B, we can use the quadratic formula. The quadratic equation in this case is x^2 + 2mx + (m^2 - 2m + 3) = 0. Comparing this equation with the standard form ax^2 + bx + c = 0, we have a = 1, b = 2m, and c = m^2 - 2m + 3.
1. Finding a+B and aB:
Using the quadratic formula, the roots of the equation are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, substituting the values of a, b, and c, we have:
x = (-(2m) ± √((2m)^2 - 4(1)(m^2 - 2m + 3))) / (2(1))
Simplifying further:
x = (-2m ± √(4m^2 - 4(m^2 - 2m + 3))) / 2
x = (-2m ± √(4m^2 - 4m^2 + 8m - 12)) / 2
x = (-2m ± √(8m - 12)) / 2
x = -m ± √(2m - 3)
Since a < B, we take a = -m - √(2m - 3) and B = -m + √(2m - 3).
To find a + B, we add the values of a and B:
a + B = (-m - √(2m - 3)) + (-m + √(2m - 3))
a + B = -2m
To find aB, we multiply the values of a and B:
aB = (-m - √(2m - 3)) * (-m + √(2m - 3))
aB = m^2 - (2m - 3)
aB = m^2 - 2m + 3
2. Finding (a - B)^2:
(a - B)^2 = (-m - √(2m - 3) - (-m + √(2m - 3)))^2
(a - B)^2 = (2√(2m - 3))^2
(a - B)^2 = 4(2m - 3)
(a - B)^2 = 8m - 12
3. If B - a = 2, we can set up the equation:
B - a = (-m + √(2m - 3)) - (-m - √(2m - 3))
2 = 2√(2m - 3)
To isolate the term with the square root, we square both sides of the equation:
4 = 4(2m - 3)
4 = 8m - 12
Simplifying further:
8m = 16
m = 2
Therefore, if B - a = 2, then m = 2.