Find a quadratic polynomial whose zeroes are 3,-2.
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Hello
If the roots are a and b, the equation of the function might be (x-a)(x-b) or most generally c(x-a)(x-b)
so you can have
[x-(2+√3)]*[x-(2-√3)]
= (x-2-√3)(x-2+√3)
= x² -2x +x√3 -2x +4 -2√3 -x√3 +2√3 -(√3)²
= x² - 4x + 4 - 3
= x² - 4x + 1
Check :
Δ = b² - 4ac = 16 - 4(1)(1) = 12
√Δ = √12 = 2√3
x1 = (4 + 2√3)/2 = 2 + √3
x2 = (4 - 2√3)/2 = 2 - √3
Hope it helps
If the roots are a and b, the equation of the function might be (x-a)(x-b) or most generally c(x-a)(x-b)
so you can have
[x-(2+√3)]*[x-(2-√3)]
= (x-2-√3)(x-2+√3)
= x² -2x +x√3 -2x +4 -2√3 -x√3 +2√3 -(√3)²
= x² - 4x + 4 - 3
= x² - 4x + 1
Check :
Δ = b² - 4ac = 16 - 4(1)(1) = 12
√Δ = √12 = 2√3
x1 = (4 + 2√3)/2 = 2 + √3
x2 = (4 - 2√3)/2 = 2 - √3
Hope it helps
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