Math help 3, Please?

Question

Prove that x^4-4x-1=0 has two different real roots?

Please take your time and consult with others, i really need the answer within 24 Hrs. This is for my little brother.

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Answer 1

By calculus you can show the function has a minimum value at x= 1 of -4. This means that the graph of the function must cross the x=axis at two points. It can't cross at just one point because then there would be three imaginary solutions which is impossible as imaginary roots always come in pairs.

In fact one of the points lies between -2.34 and x = - .265 and the other real root lies between x = 1.64 and 1.68

You can use the Newton's method to get the roots to any degree of accuracy you want. This method says that if a1 is a sufficiently close approximation to a real root of f(x)= 0 the a2= a1- f(a1)/f'(a1) gives an even closer approximation.

To fnd the miimum, take the derivative of x^4-4x-1 and set it =0 This gives 4x^3 -4 = 0
x^3-1 =0 x^3 =1 x=1
f(1) = -4