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!

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- May 5th 2011, 07:16 AMandyboy179solve the equation
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! - May 5th 2011, 09:17 AMSpringFan25
2.7 is not a solution to the equation you posted. did you type it correctly?

**edit**incorrect comment removed, oops. - May 5th 2011, 09:17 AMSambit
Let $\displaystyle f(x)=x^3-5x-8$ so that you have to find a value of $\displaystyle x$ that satisfies the equation $\displaystyle f(x)=0$. Observe that $\displaystyle f(2)<0$ and $\displaystyle f(3)>0$. So there lies x , (where $\displaystyle 2<x<3$,) such that f(x)=0.

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

After this take $\displaystyle x=2.6,2.65,...$ so on and notice when $\displaystyle f(x)$ 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 $\displaystyle 2.8025$ - May 5th 2011, 12:31 PMandyboy179
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? - May 5th 2011, 12:35 PMpoirot
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.

- May 5th 2011, 12:40 PMandyboy179
- May 5th 2011, 09:03 PMSambit
To begin with, think in simple way. $\displaystyle x=2$ and $\displaystyle x-2=0$ are equivalent. Aren't they? So here solving $\displaystyle x^3-5x=8$ is equivalent to solving $\displaystyle x^3-5x-8=0$. Alright? You are just subtracting $\displaystyle 8$ 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 $\displaystyle f(x)$ (for example, here you have $\displaystyle f(x)=x^3-5x=8$), plot $\displaystyle f(x)$ in a graph ( that is, put values of $\displaystyle x$ along horizontal X-axis and values of $\displaystyle f(x)$ along vertical Y-axis). The function will have a solution at $\displaystyle x_0$ if $\displaystyle f(x)$is $\displaystyle 0$ at $\displaystyle x=x_0$ (that is $\displaystyle f(x_0)=0$). In this case, your function looks like THIS. The blue line denotes $\displaystyle f(x)$. Notice that as $\displaystyle x$ increases, $\displaystyle f(x)$ increases ( $\displaystyle f(x)$ 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 $\displaystyle x_1 , x_2$ such that $\displaystyle f(x_1)<0$ and $\displaystyle f(x_2)>0$, because your required solution always lies in between $\displaystyle x_1$ and$\displaystyle x_2$

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

The red point is the solution: 2.80259