What are the solutions to the equation (3x-1)^2-(x+5)^2=0 ?

The solutions to the equation (3x-1)^2-(x+5)^2=0 are x=3,x=-1

Find the zeros of (3x-1)^2-(x+5)^2=0

The zeros of (3x-1)^2-(x+5)^2=0 are x=3,x=-1

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

x = 3, x = - 1

Solution steps

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

Expand

(3 x - 1)^{2} - (x + 5)^{2} : 8 x^{2} - 16 x - 24

8 x^{2} - 16 x - 24 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 8 ( - 24 )}{2 \cdot 8}

(- 16)^{2} - 4 \cdot 8 (- 24) = 32

x_{1, 2} = \frac{- ( - 16 ) \pm 32}{2 \cdot 8}

Separate the solutions

x_{1} = \frac{- ( - 16 ) + 32}{2 \cdot 8}, x_{2} = \frac{- ( - 16 ) - 32}{2 \cdot 8}

x = \frac{- ( - 16 ) + 32}{2 \cdot 8} : 3

x = \frac{- ( - 16 ) - 32}{2 \cdot 8} : - 1

The solutions to the quadratic equation are:

x = 3, x = - 1

c, A = \frac{1}{3} b c^{2}

4 x^{2} - 4 x + 15 = 0

5 x^{2} - 100 x = - 405

- 3 y^{2} - y + 2 = 0

0 = 16 - 0.01 s^{2}

What are the solutions to the equation (3x-1)^2-(x+5)^2=0 ?
The solutions to the equation (3x-1)^2-(x+5)^2=0 are x=3,x=-1
Find the zeros of (3x-1)^2-(x+5)^2=0
The zeros of (3x-1)^2-(x+5)^2=0 are x=3,x=-1