Question

I need help with math again 😭

Answer 1

Consider the equation (2x^3 - 4x^2 + 6x - 8 = 0). Find all complex solutions for (x) that satisfy this equation.

Answer 2

do u slove all of it at once our just one part first

Answer 3

surreeee what.

Answer 4

The line on my paper is like long so the answer looks like it's larger than just 1.6 or 1.7

Answer 5

Hmm. Thinking about it. Give me a second.

Answer 6

It's really complex. Here.

Answer 7

Oh my god who could've figured that out

Answer 8

If you're doing this in the context of rational roots theorem, you probably have a typo in the problem (either you copied it down wrong or your teacher/problem bank has a typo).

Typically the K12 sequence to solve this problem is to use the rational roots theorem to determine if there are any rational roots by looking at the quotient of the last and first term (a result of galois field theory). After finding that there is at least one rational root, you factor this from the cubic to obtain a quadratic equation. This quadratic equation can then be solved with the quadratic equation.

If you're doing this in the context of just solving random cubics, you can use Cardano's formula, although you might hit a roadblock in simplifying nested roots solutions. You can use what people here have given you, which are either numeric solutions to the problem (likely via Newton's method), or just throwing it at a computer algebra system (wolfram alpha, mathematica, matlab, maple, mathcad, etc) and telling it to find your solution.

But yeah, it's probably a typo and you should have a rational root.

Answer 9

All I found was this: [ \int \frac{\sin^2(x)}{e^x \cdot \sqrt{1 - \cos^2(x)}} , dx ]

Answer 10

To solve the equation \(2x^3 - 4x^2 + 6x - 8 = 0\), we can use the cubic formula or factor by grouping to simplify the equation. After factoring by grouping, we get:
\[2x^2(x - 2) + 4(x - 2) = 0\]
\[(2x^2 + 4)(x - 2) = 0\]

Setting each factor equal to zero, we find:
\[2x^2 + 4 = 0 \quad \text{and} \quad x - 2 = 0\]

Solving for \(x\) in the first equation, we get:
\[2x^2 = -4\]
\[x^2 = -2\]
\[x = \pm \sqrt{-2}\]
\[x = \pm i\sqrt{2}\]

Solving for \(x\) in the second equation, we get:
\[x = 2\]

Therefore, the complex solutions for \(x\) that satisfy the equation are:
\[x = 2\]
\[x = \pm i\sqrt{2}\]