What are the solutions to the equation x^1+2x^2=8,3x^1-4x^2=4 ?

The solutions to the equation x^1+2x^2=8,3x^1-4x^2=4 are x=(-1+sqrt(65))/4 ,x=(-1-sqrt(65))/4

Find the zeros of x^1+2x^2=8,3x^1-4x^2=4

The zeros of x^1+2x^2=8,3x^1-4x^2=4 are x=(-1+sqrt(65))/4 ,x=(-1-sqrt(65))/4

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x^{1} + 2 x^{2} = 8, 3 x^{1} - 4 x^{2} = 4

x = \frac{- 1 + 65}{4}, x = \frac{- 1 - 65}{4}

Decimal Notation

x = 1.76556 \dots, x = - 2.26556 \dots

Solution steps

x^{1} + 2 x^{2} = 8

Refine

x + 2 x^{2} = 8

Move

8

to the left side

x + 2 x^{2} - 8 = 0

Write in the standard form

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

2 x^{2} + x - 8 = 0

Solve with the quadratic formula

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

1^{2} - 4 \cdot 2 (- 8) = 65

x_{1, 2} = \frac{- 1 \pm 65}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 1 + 65}{2 \cdot 2}, x_{2} = \frac{- 1 - 65}{2 \cdot 2}

x = \frac{- 1 + 65}{2 \cdot 2} : \frac{- 1 + 65}{4}

x = \frac{- 1 - 65}{2 \cdot 2} : \frac{- 1 - 65}{4}

The solutions to the quadratic equation are:

x = \frac{- 1 + 65}{4}, x = \frac{- 1 - 65}{4}

10 t^{2} - 4 t - 32 = 0

- 5 x^{2} + 3 = 2

7 p^{2} + 10 = 26

9 v^{2} + 4 = 571

6 t - 3 t^{2} = 0

What are the solutions to the equation x^1+2x^2=8,3x^1-4x^2=4 ?
The solutions to the equation x^1+2x^2=8,3x^1-4x^2=4 are x=(-1+sqrt(65))/4 ,x=(-1-sqrt(65))/4
Find the zeros of x^1+2x^2=8,3x^1-4x^2=4
The zeros of x^1+2x^2=8,3x^1-4x^2=4 are x=(-1+sqrt(65))/4 ,x=(-1-sqrt(65))/4