What are the solutions to the equation (5x^2)/3 =3((x^2)/2-x/4) ?

The solutions to the equation (5x^2)/3 =3((x^2)/2-x/4) are x=-9/2 ,x=0

Find the zeros of (5x^2)/3 =3((x^2)/2-x/4)

The zeros of (5x^2)/3 =3((x^2)/2-x/4) are x=-9/2 ,x=0

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\frac{5 x ^{2}}{3} = 3 (\frac{x ^{2}}{2} - \frac{x}{4})

x = - \frac{9}{2}, x = 0

Solution steps

\frac{5 x ^{2}}{3} = 3 (\frac{x ^{2}}{2} - \frac{x}{4})

Multiply both sides by

3

5 x^{2} = 9 (\frac{x ^{2}}{2} - \frac{x}{4})

Expand

9 (\frac{x ^{2}}{2} - \frac{x}{4}) : \frac{9 x ^{2}}{2} - \frac{9 x}{4}

5 x^{2} = \frac{9 x ^{2}}{2} - \frac{9 x}{4}

Switch sides

\frac{9 x ^{2}}{2} - \frac{9 x}{4} = 5 x^{2}

Move

5 x^{2}

to the left side

\frac{- 2 x ^{2} - 9 x}{4} = 0

Write in the standard form

a x^{2} + b x + c = 0

- \frac{x ^{2}}{2} - \frac{9 x}{4} = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - \frac{9}{4} ) \pm ( - \frac{9}{4} ) ^{2} - 4 ( - \frac{1}{2} ) \cdot 0}{2 ( - \frac{1}{2} )}

(- \frac{9}{4})^{2} - 4 (- \frac{1}{2}) \cdot 0 = \frac{9}{4}

x_{1, 2} = \frac{- ( - \frac{9}{4} ) \pm \frac{9}{4}}{2 ( - \frac{1}{2} )}

Separate the solutions

x_{1} = \frac{- ( - \frac{9}{4} ) + \frac{9}{4}}{2 ( - \frac{1}{2} )}, x_{2} = \frac{- ( - \frac{9}{4} ) - \frac{9}{4}}{2 ( - \frac{1}{2} )}

x = \frac{- ( - \frac{9}{4} ) + \frac{9}{4}}{2 ( - \frac{1}{2} )} : - \frac{9}{2}

x = \frac{- ( - \frac{9}{4} ) - \frac{9}{4}}{2 ( - \frac{1}{2} )} : 0

The solutions to the quadratic equation are:

x = - \frac{9}{2}, x = 0

x - 1 + y - 1 z - 1

c^{2} + c 3 = 6

4 (- \frac{3}{2}) + \frac{1}{10} - (- \frac{3}{2}) (\frac{2}{5})

- 10 x^{5} - 5 x

3 (2 b - 6) + 2 = - 52

What are the solutions to the equation (5x^2)/3 =3((x^2)/2-x/4) ?
The solutions to the equation (5x^2)/3 =3((x^2)/2-x/4) are x=-9/2 ,x=0
Find the zeros of (5x^2)/3 =3((x^2)/2-x/4)
The zeros of (5x^2)/3 =3((x^2)/2-x/4) are x=-9/2 ,x=0