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

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

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

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

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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

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)

3 (x - 2) = 2 (x - 3)

\frac{( 2 n ) !}{( 2 n + 2 ) !}

1 + \frac{5}{4}

2 (x + 2) = 6 x + 2 - 4 x + 2

\frac{1}{4} (x - 7) = 1 + 3 x

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