What are the solutions to the equation (x^2)/3 =-x^2+4x ?

The solutions to the equation (x^2)/3 =-x^2+4x are x=0,x=3

Find the zeros of (x^2)/3 =-x^2+4x

The zeros of (x^2)/3 =-x^2+4x are x=0,x=3

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\frac{x ^{2}}{3} = - x^{2} + 4 x

x = 0, x = 3

Solution steps

\frac{x ^{2}}{3} = - x^{2} + 4 x

Multiply both sides by

3

x^{2} = - 3 x^{2} + 12 x

Switch sides

- 3 x^{2} + 12 x = x^{2}

Move

x^{2}

to the left side

- 4 x^{2} + 12 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 ( - 4 ) \cdot 0}{2 ( - 4 )}

1 2^{2} - 4 (- 4) \cdot 0 = 12

x_{1, 2} = \frac{- 12 \pm 12}{2 ( - 4 )}

Separate the solutions

x_{1} = \frac{- 12 + 12}{2 ( - 4 )}, x_{2} = \frac{- 12 - 12}{2 ( - 4 )}

x = \frac{- 12 + 12}{2 ( - 4 )} : 0

x = \frac{- 12 - 12}{2 ( - 4 )} : 3

The solutions to the quadratic equation are:

x = 0, x = 3

x^{2} + 2 x - 60 = 0

2.4 + t + 0.1 t^{2} = 0

5 (x^{2} - 1) = 4 (x^{2} + 1)

48 x^{2} + 12 x - 90 = 0

x^{2} - 4 x + 2 = 10

What are the solutions to the equation (x^2)/3 =-x^2+4x ?
The solutions to the equation (x^2)/3 =-x^2+4x are x=0,x=3
Find the zeros of (x^2)/3 =-x^2+4x
The zeros of (x^2)/3 =-x^2+4x are x=0,x=3