2x^3 + x^2 - 5= 0
x=?
Is this possible to solve?
Yea. x=1.209355. You have to find the roots.
Thank you, but how you made that? What was the process?
There is no "nice" formula for solving cubics.
We can graph y = 2x^3 + x^2 - 5 and get an approximate answer form the graph
http://www.wolframalpha.com/input/?i=plot+y+%3D+2x%5E3+%2B+x%5E2+-+5
and "guess" at appr 1.2
You can nibble away at the answer using a calculator
x = 1, y = 2+1-5 = -2 , too low
x = 1.2 , y = -.0104 , a bit too low but close
x = 1.21, y = .00722 , a bit too high but very close
x = 1.205 , y = -.048 , a bit too low but close
so x is between 1.205 and 1.21
you get the idea ?
If you know Calculus, use Newton's method.
let me know if you study Calculus before I show you that method
Yes, it is possible to solve the equation 2x^3 + x^2 - 5 = 0. This is a cubic equation, which means it involves a variable raised to the power of 3. To solve it, you can use various methods such as factoring, synthetic division, or the Rational Root Theorem.
One common approach to solving a cubic equation is to use the Rational Root Theorem. According to the theorem, if a rational number p/q is a root of a polynomial equation, then p must be a factor of the constant term (in this case, -5) and q must be a factor of the leading coefficient (in this case, 2).
First, find the factors of the constant term (-5). The factors of -5 are -5, -1, 1, and 5.
Next, find the factors of the leading coefficient (2). The factors of 2 are 2 and 1.
Now, create all possible combinations of the factors. These will be the potential rational roots of the equation. In this case, the potential rational roots are ±1/2, ±1, ±5/2, and ±5.
To determine if any of these potential roots are actual solutions, you can use synthetic division or substitute each potential root into the equation to see if it equals zero.
After testing the potential roots, you will find that one or more of them satisfy the equation and hence provide the values of x.