solve the equation

**andyboy179** · May 5th 2011, 07:16 AM

hi,
i nned to solve this equation:
x^3-5x=8

x=2.7

im unsure how to work this out.
i know i would do 2.7^3-13.5x=8 but i don't know what to do after this.

some help would be great!

thanks!

**Sambit** · May 5th 2011, 09:17 AM

Let $f(x)=x^3-5x-8$ so that you have to find a value of $x$ that satisfies the equation $f(x)=0$ . Observe that $f(2)<0$ and $f(3)>0$ . So there lies x , (where $2<x<3$ ,) such that f(x)=0.

See that $f(2.5) < 0$ . So for $f(x)$ to be $0$ , you have $2.5<x<3$ .

After this take $x=2.6,2.65,...$ so on and notice when $f(x)$ changes its sign. After afew couple of steps you will get the solution according to what precision you want to maintain.

However, the correct solution is $2.8025$

**SpringFan25** · May 5th 2011, 09:17 AM

2.7 is not a solution to the equation you posted. did you type it correctly?

editincorrect comment removed, oops.

**andyboy179** · May 5th 2011, 12:31 PM

springfan25- yes i have to find if 2.7 works or not and the second part is 2.8 but i was going to do that on my own.
sambit- i don't really understand what you have written. could i -8 from each side to get 2.7^3-13.5x-8=0 and do a quadratic equation?

**poirot** · May 5th 2011, 12:35 PM

This is a cubic equation so no you could not 'do a quadratic equation'. I assume you are doing numerical methods. In which case the bisection method outlined by Sambit could be used.

**andyboy179** · May 5th 2011, 12:40 PM

Originally Posted by poirot

This is a cubic equation so no you could not 'do a quadratic equation'. I assume you are doing numerical methods. In which case the bisection method outlined by Sambit could be used.

oh ok, would you care to explain it to me more simple please as i can't understand what sambit has posted.

**Sambit** · May 5th 2011, 09:03 PM

Originally Posted by andyboy179

sambit- i don't really understand what you have written. could i -8 from each side to get 2.7^3-13.5x-8=0 and do a quadratic equation?

Originally Posted by andyboy179

oh ok, would you care to explain it to me more simple please as i can't understand what sambit has posted.

To begin with, think in simple way. $x=2$ and $x-2=0$ are equivalent. Aren't they? So here solving $x^3-5x=8$ is equivalent to solving $x^3-5x-8=0$ . Alright? You are just subtracting $8$ from both the sides; so they must be equivalent.

Next try to understand what does a solution of an equation mean. I suppose you have the preliminary concept regarding function and graph. If you are gievn a function $f(x)$ (for example, here you have $f(x)=x^3-5x=8$ ), plot $f(x)$ in a graph ( that is, put values of $x$ along horizontal X-axis and values of $f(x)$ along vertical Y-axis). The function will have a solution at $x_0$ if $f(x)$ is $0$ at $x=x_0$ (that is $f(x_0)=0$ ). In this case, your function looks like THIS. The blue line denotes $f(x)$ . Notice that as $x$ increases, $f(x)$ increases ( $f(x)$ becomes +ve from -ve). During this increase, the curve cuts the X-axis at some point (red point)-- that particular point is the solution of your equation.

So this is the concept. This is why you should find two consecutive values of $x_1 , x_2$ such that $f(x_1)<0$ and $f(x_2)>0$ , because your required solution always lies in between $x_1$ and $x_2$

Now I hope you'll understand what my previous post meant.

The red point is the solution: 2.80259

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