Asked by justme
                Dtermine an equation, in simplified form, for the family of cubic functions with zeros 2 and 4+-sqrt. 3.
            
            
        Answers
                    Answered by
            Steve
            
    y = (x-2)(x-4-√3)(x-4+√3)
= (x-2)(x<sup>2</sup> - 8x + 13)
How did I get that quadratic? One way is to just multiply all the stuff out, the other is to realize that
4 ± √3 = (8 ± 2√3)/2 = (8 ± √12)
So, -b = 8
b<sup>2</sup> - 4ac = 64 - 4ac = 12
so, 4ac = 52 = 4*13
We have a=1, so c=13
y = (x-2)(x<sup>2</sup> - 8x + 13)
x<sup>3</sup> - 10x<sup>2</sup> + 29x - 26
    
= (x-2)(x<sup>2</sup> - 8x + 13)
How did I get that quadratic? One way is to just multiply all the stuff out, the other is to realize that
4 ± √3 = (8 ± 2√3)/2 = (8 ± √12)
So, -b = 8
b<sup>2</sup> - 4ac = 64 - 4ac = 12
so, 4ac = 52 = 4*13
We have a=1, so c=13
y = (x-2)(x<sup>2</sup> - 8x + 13)
x<sup>3</sup> - 10x<sup>2</sup> + 29x - 26
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