What are the solutions to the equation x^2+(-2-(sqrt(3))/3 x)^2=12 ?

The solutions to the equation x^2+(-2-(sqrt(3))/3 x)^2=12 are x=sqrt(3),x=-2sqrt(3)

Find the zeros of x^2+(-2-(sqrt(3))/3 x)^2=12

The zeros of x^2+(-2-(sqrt(3))/3 x)^2=12 are x=sqrt(3),x=-2sqrt(3)

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x^{2} + (- 2 - \frac{3}{3} x)^{2} = 12

x = 3, x = - 23

Decimal Notation

x = 1.73205 \dots, x = - 3.46410 \dots

Solution steps

x^{2} + (- 2 - \frac{3}{3} x)^{2} = 12

Expand

x^{2} + (- 2 - \frac{3}{3} x)^{2} : \frac{4 x}{3} + \frac{4 x ^{2}}{3} + 4

\frac{4 x}{3} + \frac{4 x ^{2}}{3} + 4 = 12

Move

12

to the left side

\frac{4 x ^{2}}{3} + \frac{4 x}{3} - 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- \frac{4}{3} \pm ( \frac{4}{3} ) ^{2} - 4 \cdot \frac{4}{3} ( - 8 )}{2 \cdot \frac{4}{3}}

(\frac{4}{3})^{2} - 4 \cdot \frac{4}{3} (- 8) = 43

x_{1, 2} = \frac{- \frac{4}{3} \pm 4 3}{2 \cdot \frac{4}{3}}

Separate the solutions

x_{1} = \frac{- \frac{4}{3} + 4 3}{2 \cdot \frac{4}{3}}, x_{2} = \frac{- \frac{4}{3} - 4 3}{2 \cdot \frac{4}{3}}

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

x = \frac{- \frac{4}{3} - 4 3}{2 \frac{4}{3}} : - 23

The solutions to the quadratic equation are:

x = 3, x = - 23

20 r + 3 = 103

x^{2} - 36 x

9 ∣ 9 - 8 x ∣ = 2 x + 3

7 (4 - x) = 91

lo g_{11} (1)

What are the solutions to the equation x^2+(-2-(sqrt(3))/3 x)^2=12 ?
The solutions to the equation x^2+(-2-(sqrt(3))/3 x)^2=12 are x=sqrt(3),x=-2sqrt(3)
Find the zeros of x^2+(-2-(sqrt(3))/3 x)^2=12
The zeros of x^2+(-2-(sqrt(3))/3 x)^2=12 are x=sqrt(3),x=-2sqrt(3)