1. ## Solve for x

(x-2)(x^2 + 3x) + 5x - 3 = 0

Any suggestions on how to solve for x? Thanks,

Jim

2. Originally Posted by Jim Newt
(x-2)(x^2 + 3x) + 5x - 3 = 0

Any suggestions on how to solve for x? Thanks,

Jim
0. Graph the funktion to get the number of x-intercepts.

1. Expand the brackets:

$x^3+x^2-x-3=0$

2. Use Cardano's formula (kind of tricky) or

3. Use Newton-Raphson method to get an approximate solution.

4. There is only one real solution.

5. For your confirmation only: $x \approx 1.359304085$

3. Simplifies to $x^3 + x^2 - x - 3 = 0$

No rational roots.

One irrational root
Two complex roots

4. Thanks!

5. Originally Posted by earboth
0. Graph the funktion to get the number of x-intercepts.

1. Expand the brackets:

$x^3+x^2-x-3=0$

2. Use Cardano's formula (kind of tricky) or

3. Use Newton-Raphson method to get an approximate solution.

4. There is only one real solution.

5. For your confirmation only: $x \approx 1.359304085$
Use a combination of intuitiveness and the intermediate value theorem to get a good guess to start Newton's method