What are the solutions to the equation A=(2x+3)^2-(2x-3)^2+(3x-4)^2-8x^2-16 ?

The solutions to the equation A=(2x+3)^2-(2x-3)^2+(3x-4)^2-8x^2-16 are x=sqrt(A),x=-sqrt(A)

Find the zeros of A=(2x+3)^2-(2x-3)^2+(3x-4)^2-8x^2-16

The zeros of A=(2x+3)^2-(2x-3)^2+(3x-4)^2-8x^2-16 are x=sqrt(A),x=-sqrt(A)

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A = (2 x + 3)^{2} - (2 x - 3)^{2} + (3 x - 4)^{2} - 8 x^{2} - 16

x = A, x = - A

Solution steps

A = (2 x + 3)^{2} - (2 x - 3)^{2} + (3 x - 4)^{2} - 8 x^{2} - 16

Expand

(2 x + 3)^{2} - (2 x - 3)^{2} + (3 x - 4)^{2} - 8 x^{2} - 16 : x^{2}

A = x^{2}

Switch sides

x^{2} = A

Move

A

to the left side

x^{2} - A = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 \cdot 1 \cdot ( - A )}{2 \cdot 1}

Simplify

0^{2} - 4 \cdot 1 \cdot (- A) : 2 A

x_{1, 2} = \frac{- 0 \pm 2 A}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 0 + 2 A}{2 \cdot 1}, x_{2} = \frac{- 0 - 2 A}{2 \cdot 1}

x = \frac{- 0 + 2 A}{2 \cdot 1} : A

x = \frac{- 0 - 2 A}{2 \cdot 1} : - A

The solutions to the quadratic equation are:

x = A, x = - A

(5 - x) (x + 2) = 2 - 3 (2 - x)

x^{2} + 6 x - 41 = 6 x - 5

7 x^{2} - 5 x - 16 = 0

x^{2} + 2 x - 16 = 2 x

- 2 a^{2} + 72 a = 0

What are the solutions to the equation A=(2x+3)^2-(2x-3)^2+(3x-4)^2-8x^2-16 ?
The solutions to the equation A=(2x+3)^2-(2x-3)^2+(3x-4)^2-8x^2-16 are x=sqrt(A),x=-sqrt(A)
Find the zeros of A=(2x+3)^2-(2x-3)^2+(3x-4)^2-8x^2-16
The zeros of A=(2x+3)^2-(2x-3)^2+(3x-4)^2-8x^2-16 are x=sqrt(A),x=-sqrt(A)