What are the solutions to the equation (t)^2+(3t)^2=4 ?

The solutions to the equation (t)^2+(3t)^2=4 are t=sqrt(2/5),t=-sqrt(2/5)

Find the zeros of (t)^2+(3t)^2=4

The zeros of (t)^2+(3t)^2=4 are t=sqrt(2/5),t=-sqrt(2/5)

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(t)^{2} + (3 t)^{2} = 4

t = \frac{2}{5}, t = - \frac{2}{5}

Decimal Notation

t = 0.63245 \dots, t = - 0.63245 \dots

Solution steps

t^{2} + (3 t)^{2} = 4

Expand

t^{2} + (3 t)^{2} : 10 t^{2}

10 t^{2} = 4

Move

4

to the left side

10 t^{2} - 4 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 \cdot 10 ( - 4 )}{2 \cdot 10}

0^{2} - 4 \cdot 10 (- 4) = 410

t_{1, 2} = \frac{- 0 \pm 4 10}{2 \cdot 10}

Separate the solutions

t_{1} = \frac{- 0 + 4 10}{2 \cdot 10}, t_{2} = \frac{- 0 - 4 10}{2 \cdot 10}

t = \frac{- 0 + 4 10}{2 \cdot 10} : \frac{2}{5}

t = \frac{- 0 - 4 10}{2 \cdot 10} : - \frac{2}{5}

The solutions to the quadratic equation are:

t = \frac{2}{5}, t = - \frac{2}{5}

5 t + 1 > 3 t - 13

4 t^{2} + 13 t + 10

12 \geq 6 (12 x + 2)

- 5 y^{2} + 3 y + 14

x^{2} + 1 3^{2} = 2 5^{2}

What are the solutions to the equation (t)^2+(3t)^2=4 ?
The solutions to the equation (t)^2+(3t)^2=4 are t=sqrt(2/5),t=-sqrt(2/5)
Find the zeros of (t)^2+(3t)^2=4
The zeros of (t)^2+(3t)^2=4 are t=sqrt(2/5),t=-sqrt(2/5)