What are the solutions to the equation 5/3 x+(2x+1)(1/3 x-1)=0 ?

The solutions to the equation 5/3 x+(2x+1)(1/3 x-1)=0 are x=sqrt(3/2),x=-sqrt(3/2)

Find the zeros of 5/3 x+(2x+1)(1/3 x-1)=0

The zeros of 5/3 x+(2x+1)(1/3 x-1)=0 are x=sqrt(3/2),x=-sqrt(3/2)

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\frac{5}{3} x + (2 x + 1) (\frac{1}{3} x - 1) = 0

x = \frac{3}{2}, x = - \frac{3}{2}

Solution steps

\frac{5}{3} x + (2 x + 1) (\frac{1}{3} x - 1) = 0

Multiply both sides by

3

5 x + 3 (2 x + 1) (\frac{1}{3} x - 1) = 0

Expand

5 x + 3 (2 x + 1) (\frac{1}{3} x - 1) : 2 x^{2} - 3

2 x^{2} - 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 \cdot 2 ( - 3 )}{2 \cdot 2}

0^{2} - 4 \cdot 2 (- 3) = 26

x_{1, 2} = \frac{- 0 \pm 2 6}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 0 + 2 6}{2 \cdot 2}, x_{2} = \frac{- 0 - 2 6}{2 \cdot 2}

x = \frac{- 0 + 2 6}{2 \cdot 2} : \frac{3}{2}

x = \frac{- 0 - 2 6}{2 \cdot 2} : - \frac{3}{2}

The solutions to the quadratic equation are:

x = \frac{3}{2}, x = - \frac{3}{2}

6 (x - 3) = 3 x + 9

(3 t - 1) + 7 = 8 t - (3 - 2 t)

\frac{x - 1}{x ^{2} - 6 x - 7} > 0

x^{2} - 9 x + 21 = 0

\frac{x - 1}{x} < 4

What are the solutions to the equation 5/3 x+(2x+1)(1/3 x-1)=0 ?
The solutions to the equation 5/3 x+(2x+1)(1/3 x-1)=0 are x=sqrt(3/2),x=-sqrt(3/2)
Find the zeros of 5/3 x+(2x+1)(1/3 x-1)=0
The zeros of 5/3 x+(2x+1)(1/3 x-1)=0 are x=sqrt(3/2),x=-sqrt(3/2)