What are the solutions to the equation (2x-1)^2-(x+3)^2=0 ?

The solutions to the equation (2x-1)^2-(x+3)^2=0 are x=4,x=-2/3

Find the zeros of (2x-1)^2-(x+3)^2=0

The zeros of (2x-1)^2-(x+3)^2=0 are x=4,x=-2/3

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(2 x - 1)^{2} - (x + 3)^{2} = 0

x = 4, x = - \frac{2}{3}

Decimal Notation

x = 4, x = - 0.66666 \dots

Solution steps

(2 x - 1)^{2} - (x + 3)^{2} = 0

Expand

(2 x - 1)^{2} - (x + 3)^{2} : 3 x^{2} - 10 x - 8

3 x^{2} - 10 x - 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 3 ( - 8 )}{2 \cdot 3}

(- 10)^{2} - 4 \cdot 3 (- 8) = 14

x_{1, 2} = \frac{- ( - 10 ) \pm 14}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 10 ) + 14}{2 \cdot 3}, x_{2} = \frac{- ( - 10 ) - 14}{2 \cdot 3}

x = \frac{- ( - 10 ) + 14}{2 \cdot 3} : 4

x = \frac{- ( - 10 ) - 14}{2 \cdot 3} : - \frac{2}{3}

The solutions to the quadratic equation are:

x = 4, x = - \frac{2}{3}

(x^{2} + 1) 2 - 5 x^{2} - 5 = 0

x^{2} - 12 a x + 32 a^{2} = 0

- 5 x - 4 = - 6 x^{2}

y^{2} - 2 y = 8

2 x^{2} + 20 x + 48 = 0

What are the solutions to the equation (2x-1)^2-(x+3)^2=0 ?
The solutions to the equation (2x-1)^2-(x+3)^2=0 are x=4,x=-2/3
Find the zeros of (2x-1)^2-(x+3)^2=0
The zeros of (2x-1)^2-(x+3)^2=0 are x=4,x=-2/3