The logical mates

Algebra Level 2

Find the real solution of equation x 3 − x = 1 x^3-x=1


The answer is 1.34271795724475.

Hardik Chandak by Hardik Chandak , 25, India

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2 solutions

Let's solve it by using Cardano's formula:

For x x in x 3 + p x + q = 0 x^{3}+px+q=0 , the first (and posibility the only one) real solution is:

x = − q 2 + ( q 2 ) 2 + ( p 3 ) 3 3 + − q 2 − ( q 2 ) 2 + ( p 3 ) 3 3 x=\sqrt[3]{-\dfrac{q}{2}+\sqrt{(\dfrac{q}{2})^{2}+(\dfrac{p}{3})^{3}}}+\sqrt[3]{-\dfrac{q}{2}-\sqrt{(\dfrac{q}{2})^{2}+(\dfrac{p}{3})^{3}}}

In the given equation, p = − 1 p=-1 and q = − 1 q=-1 , so the real solution is:

x = 9 + 69 18 3 + 9 − 69 18 3 x=\sqrt[3]{\dfrac{9+\sqrt{69}}{18}}+\sqrt[3]{\dfrac{9-\sqrt{69}}{18}} , which is approximately x = 1.32471 x=\boxed{1.32471}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

let,f(x)=x^3-x-1 for: x=1,f(x)<0 for: x=2,f(x)>0 so root lies between x=1 and x=2 for:x=1.3,f(x)<0 for:x=1.5,f(x)>0 so now the root lies between x=1.3 and x=1.5 now on further reducing the interval length on x-axis we get the root as 1.3275

2 Helpful 0 Interesting 0 Brilliant 0 Confused

This is not a real solution .It is the hypothesis of solution.If you want to solve it algebraically you need to solve it algebraic equation.When we consider x^3 for then it x contains 3 values,what are the other 2 values.You can not deny other 2 value.Other 2 value have to prove the equation.

Imam Hossain Rony - 7 years, 2 months ago

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