What are the solutions to the equation 8n^3-4n^2=18n-9 ?

The solutions to the equation 8n^3-4n^2=18n-9 are n= 1/2 ,n=-3/2 ,n= 3/2

Find the zeros of 8n^3-4n^2=18n-9

The zeros of 8n^3-4n^2=18n-9 are n= 1/2 ,n=-3/2 ,n= 3/2

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8 n^{3} - 4 n^{2} = 18 n - 9

n = \frac{1}{2}, n = - \frac{3}{2}, n = \frac{3}{2}

Decimal Notation

n = 0.5, n = - 1.5, n = 1.5

Solution steps

8 n^{3} - 4 n^{2} = 18 n - 9

Move

9

to the left side

8 n^{3} - 4 n^{2} + 9 = 18 n

Move

18 n

to the left side

8 n^{3} - 4 n^{2} + 9 - 18 n = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

8 n^{3} - 4 n^{2} - 18 n + 9 = 0

Factor

8 n^{3} - 4 n^{2} - 18 n + 9 : (2 n - 1) (2 n + 3) (2 n - 3)

(2 n - 1) (2 n + 3) (2 n - 3) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

2 n - 1 = 0 or 2 n + 3 = 0 or 2 n - 3 = 0

Solve

2 n - 1 = 0 : n = \frac{1}{2}

Solve

2 n + 3 = 0 : n = - \frac{3}{2}

Solve

2 n - 3 = 0 : n = \frac{3}{2}

The solutions are

n = \frac{1}{2}, n = - \frac{3}{2}, n = \frac{3}{2}

x^{3} - 8 x - 3 = 0

15.625 = s^{3}

x^{4} - 5 x^{2} + 2 x + 11 = 0

(x - 1)^{3} - 54 = 0

2 x^{2} = x^{4} - 2 x^{2}

What are the solutions to the equation 8n^3-4n^2=18n-9 ?
The solutions to the equation 8n^3-4n^2=18n-9 are n= 1/2 ,n=-3/2 ,n= 3/2
Find the zeros of 8n^3-4n^2=18n-9
The zeros of 8n^3-4n^2=18n-9 are n= 1/2 ,n=-3/2 ,n= 3/2