What are the solutions to the equation 1x^2+(2x)^2=4.3 ?

The solutions to the equation 1x^2+(2x)^2=4.3 are x=(sqrt(43))/(5sqrt(2)),x=-(sqrt(43))/(5sqrt(2))

Find the zeros of 1x^2+(2x)^2=4.3

The zeros of 1x^2+(2x)^2=4.3 are x=(sqrt(43))/(5sqrt(2)),x=-(sqrt(43))/(5sqrt(2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

1 x^{2} + (2 x)^{2} = 4.3

x = \frac{43}{5 2}, x = - \frac{43}{5 2}

Decimal Notation

x = 0.92736 \dots, x = - 0.92736 \dots

Solution steps

1 \cdot x^{2} + (2 x)^{2} = 4.3

Multiply both sides by

10

50 x^{2} = 43

Move

43

to the left side

50 x^{2} - 43 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 \cdot 50 ( - 43 )}{2 \cdot 50}

0^{2} - 4 \cdot 50 (- 43) = 1086

x_{1, 2} = \frac{- 0 \pm 10 86}{2 \cdot 50}

Separate the solutions

x_{1} = \frac{- 0 + 10 86}{2 \cdot 50}, x_{2} = \frac{- 0 - 10 86}{2 \cdot 50}

x = \frac{- 0 + 10 86}{2 \cdot 50} : \frac{43}{5 2}

x = \frac{- 0 - 10 86}{2 \cdot 50} : - \frac{43}{5 2}

The solutions to the quadratic equation are:

x = \frac{43}{5 2}, x = - \frac{43}{5 2}

x^{2} + 144 x = 0

x, x^{2} - 2 y - 1 = C

3 x (x - 3) = 8 x^{2} - 11 x

y, y^{2} = x^{2} - 1

2 x^{2} - 13 x + 7 = 0

What are the solutions to the equation 1x^2+(2x)^2=4.3 ?
The solutions to the equation 1x^2+(2x)^2=4.3 are x=(sqrt(43))/(5sqrt(2)),x=-(sqrt(43))/(5sqrt(2))
Find the zeros of 1x^2+(2x)^2=4.3
The zeros of 1x^2+(2x)^2=4.3 are x=(sqrt(43))/(5sqrt(2)),x=-(sqrt(43))/(5sqrt(2))