I took the derivative of 4m^2 - 2m -4 to get m=1/4. I showed this to my instructor and this is the reply I got.
Good observation!
then, you just consider m>=2 or m<=-1, and find the minimum value of function y=4m^2-2m-4. It is solvable
I'm not sure what he means by this. I looked at...
substitute x1 into the equation to get
x1^2 -2mx1 +m+2
Substitute x2 to get
x2^2 -2mx2 +m+2
set them = to each other and solve for m
x1^2 -2mx1 + m + 2 = x2^2 -2mx2 +m + 2
so
x1^2 - 2mx1 = x2^2 -2mx2
then
x1^2 -x2^2 = m2(x1 - x2)
then factor and divide to get
m=(x1 + x2)/2
Homework Statement
4. If x1, x2 are two real number roots of real number coefficient quadratic equation:
$$x^2 -2mx + m + 2 =0$$
Solve the following questions:
(1) What are the values of m so that x1=x2?
(2) What are the...
given a*(1-b^2)^1/2 +b(1-a^2)^1\2 =1 prove a^2 + b^2 =1
I tried squaring both sides and then squaring again to get
a^4 + b^4 -2b^2 -2a^2 +2a^2b^2 +1 =0
and that could be (a^2 + b^2)(a^2 + b^2) - 2(a^2 + b^2) = -1
I don't know where to go from there and not sure this is even correct...
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