Use quadratic formula to solve each equation 3k^2+6=6
I get 3/2 but I can't seem to get the other half of the equation such as x=3.7 or .26 etc...
i think you mean 3k^2 + 6k = 6
thus, we can re write this as
3k^2 + 6k - 6 = 0
note that we can factor out 3 from each term and cancel it. thus we can re write this as
k^2 + 2k - 2 = 0
using quadratic formula,
k = [-b +- sqrt(b^2 - 4ac)]/(2a)
substituting,
k = [-2 +- sqrt(2^2 - 4*1*(-2))]/(2*1)
k = [-2 +- sqrt(4 + 8)]/2
k = [-2 +- sqrt(12)]/2
k = [-2 +- 2*sqrt(3)]/2
k = -1 +- sqrt(3)
separating into plus and minus,
k = -1 + sqrt(3) and k = -1 - sqrt(3)
hope this helps~ :)
no.. there is no 6 in front of the k at least not in my text book
To solve the equation 3k^2 + 6 = 6 using the quadratic formula, you should start by rewriting the equation in the form ax^2 + bx + c = 0, where a, b, and c are the coefficients.
In this case, rearrange the equation to get:
3k^2 + 6 - 6 = 0
3k^2 = 0
The equation is now in the desired form, with a = 3, b = 0, and c = 0.
Next, you can apply the quadratic formula, which states that for an equation of the form ax^2 + bx + c = 0, the solutions can be found using the following formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values from our equation, we get:
k = (-0 ± √(0^2 - 4 * 3 * 0)) / (2 * 3)
Simplifying further:
k = (± √(0 - 0)) / 6
k = (± √0) / 6
Since the discriminant (b^2 - 4ac) is zero, the square root of zero is zero. Therefore, we get:
k = ± 0 / 6
k = 0
Hence, the equation 3k^2 + 6 = 6 has one real solution, which is k = 0.