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

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

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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^{2} + x^{2} = (x + 3)^{2}

x = - 3 (2 - 1), x = 3 (1 + 2)

Decimal Notation

x = - 1.24264 \dots, x = 7.24264 \dots

Solution steps

x^{2} + x^{2} = (x + 3)^{2}

Expand

(x + 3)^{2} : x^{2} + 6 x + 9

2 x^{2} = x^{2} + 6 x + 9

Switch sides

x^{2} + 6 x + 9 = 2 x^{2}

Move

2 x^{2}

to the left side

- x^{2} + 6 x + 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 ( - 1 ) \cdot 9}{2 ( - 1 )}

6^{2} - 4 (- 1) \cdot 9 = 62

x_{1, 2} = \frac{- 6 \pm 6 2}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 6 + 6 2}{2 ( - 1 )}, x_{2} = \frac{- 6 - 6 2}{2 ( - 1 )}

x = \frac{- 6 + 6 2}{2 ( - 1 )} : - 3 (2 - 1)

x = \frac{- 6 - 6 2}{2 ( - 1 )} : 3 (1 + 2)

The solutions to the quadratic equation are:

x = - 3 (2 - 1), x = 3 (1 + 2)

- \frac{16}{9}

3 m^{2} + 8 m - 3

8 a^{2} b + 16 a^{3} b - 24 a^{2} b^{2}

8 - ∣ 3 x + 2 ∣ = 9

5 + 7

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