Question

Find the real root of equation
x√3+x-1=0
.​

Answer 1

Answer:

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Step-by-step explanation:

Answer 2

Being a polynomial equation of odd degree, the equation f(x) = x^3 +x -1 = 0 has at least one real root. Now f '(x) = 3. x^2 +1 > 0 for all x. Therefore there is only one real root, because between any two real roots the derivative must have value 0 somewhere between them by Rolle's theorem.

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