What are the solutions to the equation ((x+3)(x-3)-4)/2-(x-2)/3 =((x-2)^2+1)/6 ?

The solutions to the equation ((x+3)(x-3)-4)/2-(x-2)/3 =((x-2)^2+1)/6 are x=4,x=-5

Find the zeros of ((x+3)(x-3)-4)/2-(x-2)/3 =((x-2)^2+1)/6

The zeros of ((x+3)(x-3)-4)/2-(x-2)/3 =((x-2)^2+1)/6 are x=4,x=-5

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\frac{( x + 3 ) ( x - 3 ) - 4}{2} - \frac{x - 2}{3} = \frac{( x - 2 ) ^{2} + 1}{6}

x = 4, x = - 5

Solution steps

\frac{( x + 3 ) ( x - 3 ) - 4}{2} - \frac{x - 2}{3} = \frac{( x - 2 ) ^{2} + 1}{6}

Find Least Common Multiplier of

2, 3, 6 : 6

Multiply by LCM=

6

\frac{( x + 3 ) ( x - 3 ) - 4}{2} \cdot 6 - \frac{x - 2}{3} \cdot 6 = \frac{( x - 2 ) ^{2} + 1}{6} \cdot 6

Simplify

3 ((x + 3) (x - 3) - 4) - 2 (x - 2) = (x - 2)^{2} + 1

Expand

3 ((x + 3) (x - 3) - 4) - 2 (x - 2) : 3 x^{2} - 2 x - 35

Expand

(x - 2)^{2} + 1 : x^{2} - 4 x + 5

3 x^{2} - 2 x - 35 = x^{2} - 4 x + 5

Move

5

to the left side

3 x^{2} - 2 x - 40 = x^{2} - 4 x

Move

4 x

to the left side

3 x^{2} + 2 x - 40 = x^{2}

Move

x^{2}

to the left side

2 x^{2} + 2 x - 40 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 2 ( - 40 )}{2 \cdot 2}

2^{2} - 4 \cdot 2 (- 40) = 18

x_{1, 2} = \frac{- 2 \pm 18}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 2 + 18}{2 \cdot 2}, x_{2} = \frac{- 2 - 18}{2 \cdot 2}

x = \frac{- 2 + 18}{2 \cdot 2} : 4

x = \frac{- 2 - 18}{2 \cdot 2} : - 5

The solutions to the quadratic equation are:

x = 4, x = - 5

3^{2} + 6^{2} = c^{2}

\frac{x ^{3} + 8}{x ^{4} - 16}

- 8 \leq 5 x - 7 \leq 5

\frac{π}{6} - 2 π

c^{2} - 12 c = - 76

What are the solutions to the equation ((x+3)(x-3)-4)/2-(x-2)/3 =((x-2)^2+1)/6 ?
The solutions to the equation ((x+3)(x-3)-4)/2-(x-2)/3 =((x-2)^2+1)/6 are x=4,x=-5
Find the zeros of ((x+3)(x-3)-4)/2-(x-2)/3 =((x-2)^2+1)/6
The zeros of ((x+3)(x-3)-4)/2-(x-2)/3 =((x-2)^2+1)/6 are x=4,x=-5