Anonymous
Anonymous asked in Science & MathematicsMathematics · 10 months ago

# Calculate x^2-mx+m>0?

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• I think m is a constant .

So x^2 - mx + m is a quadratic function of x , its graph becomes "U" shape .

We know the solution of x^2 - mx + m = 0 is x = (1/2)[m ± √(m^2 - 4m)] ,

so the answer for x^2 - mx + m > 0 becomes :

Case1. when m < 0

m^2 - 4m > 0 , so there are 2 solutions of x . Therefore , the answer is

x < (1/2)[m - √(m^2 - 4m)] or x > (1/2)[m + √(m^2 - 4m)] .

Case2. when m = 0

m^2 - 4m = 0 , so there is only one solution of x . Therefore , the answer is

x ≠ (1/2)m . That is , x ≠ 0 .

Case3. when 0 < m < 4

m^2 - 4m < 0 , so there is no real solution of x . Therefore , the answer is

"x can be any real number" .

Case4. when m = 4

m^2 - 4m = 0 , so there is only one solution of x . Therefore , the answer is

x ≠ (1/2)m . That is , x ≠ 2 .

Case5. when m > 4

m^2 - 4m > 0 , so there are 2 solutions of x . Therefore , the answer is

x < (1/2)[m - √(m^2 - 4m)] or x > (1/2)[m + √(m^2 - 4m)] .

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• x<1/2 (m - sqrt((m - 4) m)) and m<=0

x>1/2 (m + sqrt((m - 4) m)) and m<=0

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• calculate what? you have one inequality and two variables.

There is no solution.

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