What are the solutions to the equation (-13/3 x)^2+(-2/3 x)^2+x^2=1 ?

The solutions to the equation (-13/3 x)^2+(-2/3 x)^2+x^2=1 are x= 3/(sqrt(182)),x=-3/(sqrt(182))

Find the zeros of (-13/3 x)^2+(-2/3 x)^2+x^2=1

The zeros of (-13/3 x)^2+(-2/3 x)^2+x^2=1 are x= 3/(sqrt(182)),x=-3/(sqrt(182))

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(- \frac{13}{3} x)^{2} + (- \frac{2}{3} x)^{2} + x^{2} = 1

x = \frac{3}{182}, x = - \frac{3}{182}

Decimal Notation

x = 0.22237 \dots, x = - 0.22237 \dots

Solution steps

(- \frac{13}{3} x)^{2} + (- \frac{2}{3} x)^{2} + x^{2} = 1

Expand

(- \frac{13}{3} x)^{2} + (- \frac{2}{3} x)^{2} + x^{2} : \frac{182}{9} x^{2}

\frac{182}{9} x^{2} = 1

Move

1

to the left side

\frac{182}{9} x^{2} - 1 = 0

Write in the standard form

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

\frac{182 x ^{2}}{9} - 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 \cdot \frac{182}{9} ( - 1 )}{2 \cdot \frac{182}{9}}

0^{2} - 4 \cdot \frac{182}{9} (- 1) = \frac{2 182}{3}

x_{1, 2} = \frac{- 0 \pm \frac{2 182}{3}}{2 \cdot \frac{182}{9}}

Separate the solutions

x_{1} = \frac{- 0 + \frac{2 182}{3}}{2 \cdot \frac{182}{9}}, x_{2} = \frac{- 0 - \frac{2 182}{3}}{2 \cdot \frac{182}{9}}

x = \frac{- 0 + \frac{2 182}{3}}{2 \frac{182}{9}} : \frac{3}{182}

x = \frac{- 0 - \frac{2 182}{3}}{2 \frac{182}{9}} : - \frac{3}{182}

The solutions to the quadratic equation are:

x = \frac{3}{182}, x = - \frac{3}{182}

2 x^{2} + 6 x - 196 = 0

4 m^{2} - 8 m + 9 = 0

26 x^{2} + 41 x + 20 = 6 x^{2}

- 0.00015 x^{2} - 0.001652 x + 0.0153 = 0

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

What are the solutions to the equation (-13/3 x)^2+(-2/3 x)^2+x^2=1 ?
The solutions to the equation (-13/3 x)^2+(-2/3 x)^2+x^2=1 are x= 3/(sqrt(182)),x=-3/(sqrt(182))
Find the zeros of (-13/3 x)^2+(-2/3 x)^2+x^2=1
The zeros of (-13/3 x)^2+(-2/3 x)^2+x^2=1 are x= 3/(sqrt(182)),x=-3/(sqrt(182))