help me Use the Quadratic Formula to solve N2=6n
n^2 = n.
n^2 - n = 0,
n = (-b+-sqrt(b^2-4ac)) / 2a.
n = (1+-sqrt(1-4*1*0)0 / 2,
n = (1+-sqrt(1-0)) / 2,
n = (1+-1) / 2,
n = (1+1) / 2 = 1.
n = (1-1) / 2 = 0/2 = 0.
Solution set: n = 1, and 0.
Correction:
n^2 = 6n.
n^2 - 6n = 0,
n = (6 +- sqrt(36-4*1*0)) / 2,
n = (6 +- sqrt(36-0)) / 2,
n = (6 +-sqrt(36)) / 2,
n = (6 +- 6) / 2,
n = (6+6) / 2 = 6.
n = (6-6) / 2 = 0/6 = 0.
Solution set: n = 6, and 0.
n^2= 6n
n^2-6n=0
n(n-6)=0
n=0,6
To use the quadratic formula to solve the equation N^2 = 6n, we need to rewrite the equation in standard quadratic form, which is ax^2 + bx + c = 0.
In this case, we have N^2 - 6n = 0. Notice that the equation has N^2 as the quadratic term, -6n as the linear term, and 0 as the constant term.
By comparing this to the standard form, we can say that a = 1 (because the coefficient of N^2 is 1), b = -6 (because the coefficient of n is -6), and c = 0.
Now, we can substitute these values into the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our case, x refers to the solutions for N. So, we have:
N = (-(-6) ± √((-6)^2 - 4(1)(0))) / (2(1))
Simplifying this expression gives us:
N = (6 ± √(36)) / 2
N = (6 ± 6) / 2
N = (6 + 6) / 2 or N = (6 - 6) / 2
N = 12 / 2 or N = 0 / 2
N = 6 or N = 0
Therefore, the solutions to the equation N^2 = 6n are N = 6 and N = 0.