SOLUTION: approximate to the nearest tenth, the real root of the equation f(x)=x^3-4=0

Question 202322: approximate to the nearest tenth, the real root of the equation f(x)=x^3-4=0
Found 2 solutions by dyakobovitch, solver91311:
Answer by dyakobovitch(40) (Show Source): You can put this solution on YOUR website!
The simple method: x^3=4. Raise to the 1/3 power, and you have x=4^(/3). Take the cube root of 4, and you have your answer as the method below shows:

Solved by pluggable solver: EXPLAIN simplification of an expression

Your Result:

YOUR ANSWER

This is an equation! Solutions: x=1.5874010519682.

Graphical form: simplifies to
Text form: x^3-4=0 simplifies to +x^2+1.5874010519682*x+2.51984209978975=0
Cartoon (animation) form:
For tutors: simplify_cartoon( x^3-4=0 )

If you have a website, here's a link to this solution.

DETAILED EXPLANATION

Look at .
Divided polynomial on the left by known factors (x-(1.5874010519682))
It becomes .
Result:
This is an equation! Solutions: x=1.5874010519682.

Universal Simplifier and Solver

Done!

Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

Since

we can say

And then simply take the cube root of both sides to find the real number root. There are actually two other complex number roots, but those are not what we are looking for here.

The simple way is to punch in 4 inverse x^3 into your calculator and then round off to the nearest tenth. But it is also pretty easy to calculate on paper since you don't have to go beyond the 1st decimal place.

First of all, we know that and so we know that 2 is too big and 1 is too small. So let's try 1.5. . Closer, but still a little small, so let's try 1.6. , a touch large. So now we have it bracketed. . But we can also see that is closer to 4 than . Hence the answer, to the nearest tenth, is 1.6.

John

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