Asked by Mica
I asked this question previously, but I was told you were unable to get the answer with rational numbers. I asked my teacher, and he said we could use estimates.
y=x^4-4x^3+16
y=(x-2)(x^3-2x^2-4x-8)
so i used 3.68
and i got this (x-2)(x-3.68)(x^2+1.68x+2.1824)
Could someone help me finish?
y=x^4-4x^3+16
y=(x-2)(x^3-2x^2-4x-8)
so i used 3.68
and i got this (x-2)(x-3.68)(x^2+1.68x+2.1824)
Could someone help me finish?
Answers
Answered by
Reiny
The remaining quadratic has no solution in the set of real numbers.
The value of b^2 - 4ac in the formula is negative, so we cannot take the √ of a negative number.
your only real solutions are x = 2 and x = appr. 3.68
The value of b^2 - 4ac in the formula is negative, so we cannot take the √ of a negative number.
your only real solutions are x = 2 and x = appr. 3.68
Answered by
Reiny
When you first asked this question, you asked for the factors, you did not ask for a solution to the corresponding equation x^4-4x^3+16 = 0
That is how it was answered.
That is how it was answered.
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