What are the solutions to the equation x^2+(x+4)^2=(x+9)^2 ?

The solutions to the equation x^2+(x+4)^2=(x+9)^2 are x=5+3sqrt(10),x=5-3sqrt(10)

Find the zeros of x^2+(x+4)^2=(x+9)^2

The zeros of x^2+(x+4)^2=(x+9)^2 are x=5+3sqrt(10),x=5-3sqrt(10)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^{2} + (x + 4)^{2} = (x + 9)^{2}

x = 5 + 310, x = 5 - 310

Decimal Notation

x = 14.48683 \dots, x = - 4.48683 \dots

Solution steps

x^{2} + (x + 4)^{2} = (x + 9)^{2}

Expand

x^{2} + (x + 4)^{2} : 2 x^{2} + 8 x + 16

Expand

(x + 9)^{2} : x^{2} + 18 x + 81

2 x^{2} + 8 x + 16 = x^{2} + 18 x + 81

Move

81

to the left side

2 x^{2} + 8 x - 65 = x^{2} + 18 x

Move

18 x

to the left side

2 x^{2} - 10 x - 65 = x^{2}

Move

x^{2}

to the left side

x^{2} - 10 x - 65 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 1 \cdot ( - 65 )}{2 \cdot 1}

(- 10)^{2} - 4 \cdot 1 \cdot (- 65) = 610

x_{1, 2} = \frac{- ( - 10 ) \pm 6 10}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 10 ) + 6 10}{2 \cdot 1}, x_{2} = \frac{- ( - 10 ) - 6 10}{2 \cdot 1}

x = \frac{- ( - 10 ) + 6 10}{2 \cdot 1} : 5 + 310

x = \frac{- ( - 10 ) - 6 10}{2 \cdot 1} : 5 - 310

The solutions to the quadratic equation are:

x = 5 + 310, x = 5 - 310

0. 6^{2}

1 < ∣ x ∣

0.131 = 767.77 (x) + 0.0198

(x + 6) (x + 4)

\frac{6 - x}{3} - \frac{3 + x}{2} = 4

What are the solutions to the equation x^2+(x+4)^2=(x+9)^2 ?
The solutions to the equation x^2+(x+4)^2=(x+9)^2 are x=5+3sqrt(10),x=5-3sqrt(10)
Find the zeros of x^2+(x+4)^2=(x+9)^2
The zeros of x^2+(x+4)^2=(x+9)^2 are x=5+3sqrt(10),x=5-3sqrt(10)