What are the solutions to the equation 36^2=x^2+((2x)/3)^2 ?

The solutions to the equation 36^2=x^2+((2x)/3)^2 are x=(108)/(sqrt(13)),x=-(108)/(sqrt(13))

Find the zeros of 36^2=x^2+((2x)/3)^2

The zeros of 36^2=x^2+((2x)/3)^2 are x=(108)/(sqrt(13)),x=-(108)/(sqrt(13))

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3 6^{2} = x^{2} + (\frac{2 x}{3})^{2}

x = \frac{108}{13}, x = - \frac{108}{13}

Decimal Notation

x = 29.95381 \dots, x = - 29.95381 \dots

Solution steps

3 6^{2} = x^{2} + (\frac{2 x}{3})^{2}

Expand

3 6^{2} : 1296

Expand

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

1296 = \frac{13 x ^{2}}{9}

Switch sides

\frac{13 x ^{2}}{9} = 1296

Move

1296

to the left side

\frac{13 x ^{2}}{9} - 1296 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 \cdot \frac{13}{9} ( - 1296 )}{2 \cdot \frac{13}{9}}

0^{2} - 4 \cdot \frac{13}{9} (- 1296) = 2413

x_{1, 2} = \frac{- 0 \pm 24 13}{2 \cdot \frac{13}{9}}

Separate the solutions

x_{1} = \frac{- 0 + 24 13}{2 \cdot \frac{13}{9}}, x_{2} = \frac{- 0 - 24 13}{2 \cdot \frac{13}{9}}

x = \frac{- 0 + 24 13}{2 \frac{13}{9}} : \frac{108}{13}

x = \frac{- 0 - 24 13}{2 \frac{13}{9}} : - \frac{108}{13}

The solutions to the quadratic equation are:

x = \frac{108}{13}, x = - \frac{108}{13}

16 x^{2} + 9 = 25

(x - 5) (x - 4) = x - 9

x, x^{2} + (- k - x)^{2} = \frac{k ^{2}}{4} + 1

- x^{2} + 14 x - 48 = 0

2 r^{2} - 2 r - 15 = - 2

What are the solutions to the equation 36^2=x^2+((2x)/3)^2 ?
The solutions to the equation 36^2=x^2+((2x)/3)^2 are x=(108)/(sqrt(13)),x=-(108)/(sqrt(13))
Find the zeros of 36^2=x^2+((2x)/3)^2
The zeros of 36^2=x^2+((2x)/3)^2 are x=(108)/(sqrt(13)),x=-(108)/(sqrt(13))