Question

What is the solution to the equation 0=10k^2-8k+8 in terms of k?

Answer 1

divide both side by 2 - just to simplify.
use quadratic formula:
k=2+/- 4i

Answer 2

Use the quadratic equation to solve.

\[k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

Where....

\[0 = ak^2 - bk + c\]

a, b, and c are the constants in the equation. Substitute in and watch the magic happen. If you still need help, ask me to clarify.

Answer 3

8 +/- 16i
k = ----------
20

4(2 +/- 4i) 2 +/- 4i
k = ---------- = ---------
4*5 5

Answer 4

Possible intermediate steps:
0 = 10 k^2-8 k+8
Subtract (10 k^2-8 k+8) from both sides:
-10 k^2+8 k-8 = 0
Solve the quadratic equation by completing the square:

Divide both sides by -10:
k^2-(4 k)/5+4/5 = 0
Subtract 4/5 from both sides:
k^2-(4 k)/5 = -4/5
Add 4/25 to both sides:
k^2-(4 k)/5+4/25 = -16/25
Factor the left hand side:
(k-2/5)^2 = -16/25
Take the square root of both sides:
abs(k-2/5) = (4 i)/5
Eliminate the absolute value:
k-2/5 = -(4 i)/5 or k-2/5 = (4 i)/5
Add 2/5 to both sides:
k = 2/5-(4 i)/5 or k-2/5 = (4 i)/5
Add 2/5 to both sides:
k = 2/5-(4 i)/5 or k = 2/5+(4 i)/5

Answer 5

Oh ok i get it now thank you for explaining.