What are the solutions to the equation 9a^3-441a=-4a^2+196 ?

The solutions to the equation 9a^3-441a=-4a^2+196 are a=-4/9 ,a=-7,a=7

Find the zeros of 9a^3-441a=-4a^2+196

The zeros of 9a^3-441a=-4a^2+196 are a=-4/9 ,a=-7,a=7

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9 a^{3} - 441 a = - 4 a^{2} + 196

a = - \frac{4}{9}, a = - 7, a = 7

Decimal Notation

a = - 0.44444 \dots, a = - 7, a = 7

Solution steps

9 a^{3} - 441 a = - 4 a^{2} + 196

Move

196

to the left side

9 a^{3} - 441 a - 196 = - 4 a^{2}

Move

4 a^{2}

to the left side

9 a^{3} - 441 a - 196 + 4 a^{2} = 0

Write in the standard form

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

9 a^{3} + 4 a^{2} - 441 a - 196 = 0

Factor

9 a^{3} + 4 a^{2} - 441 a - 196 : (9 a + 4) (a + 7) (a - 7)

(9 a + 4) (a + 7) (a - 7) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

9 a + 4 = 0 or a + 7 = 0 or a - 7 = 0

Solve

9 a + 4 = 0 : a = - \frac{4}{9}

Solve

a + 7 = 0 : a = - 7

Solve

a - 7 = 0 : a = 7

The solutions are

a = - \frac{4}{9}, a = - 7, a = 7

\frac{2 - i}{1 + 2 i} \cdot z^{4} = 1

x^{3} + x^{2} + 2 x + 1 = 0

x^{3} - 40 x^{2} + 404 x - 80 = 0

(x + 2) (x + 3) (x + 8) (x + 12) = 54 x^{2}

8 x^{3} + 64 x^{2} + 120 x = 0

What are the solutions to the equation 9a^3-441a=-4a^2+196 ?
The solutions to the equation 9a^3-441a=-4a^2+196 are a=-4/9 ,a=-7,a=7
Find the zeros of 9a^3-441a=-4a^2+196
The zeros of 9a^3-441a=-4a^2+196 are a=-4/9 ,a=-7,a=7