What are the solutions to the equation (x+7}{12}-\frac{x^2+1)/4 =1-(x^2+2)/3 ?

The solutions to the equation (x+7}{12}-\frac{x^2+1)/4 =1-(x^2+2)/3 are x=0,x=-1

Find the zeros of (x+7}{12}-\frac{x^2+1)/4 =1-(x^2+2)/3

The zeros of (x+7}{12}-\frac{x^2+1)/4 =1-(x^2+2)/3 are x=0,x=-1

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\frac{x + 7}{12} - \frac{x ^{2} + 1}{4} = 1 - \frac{x ^{2} + 2}{3}

x = 0, x = - 1

Solution steps

\frac{x + 7}{12} - \frac{x ^{2} + 1}{4} = 1 - \frac{x ^{2} + 2}{3}

Find Least Common Multiplier of

12, 4, 3 : 12

Multiply by LCM=

12

\frac{x + 7}{12} \cdot 12 - \frac{x ^{2} + 1}{4} \cdot 12 = 1 \cdot 12 - \frac{x ^{2} + 2}{3} \cdot 12

Simplify

x + 7 - 3 (x^{2} + 1) = 12 - 4 (x^{2} + 2)

Expand

x + 7 - 3 (x^{2} + 1) : - 3 x^{2} + x + 4

Expand

12 - 4 (x^{2} + 2) : - 4 x^{2} + 4

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

Move

4

to the left side

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

Move

4 x^{2}

to the left side

x^{2} + x = 0

Solve with the quadratic formula

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

1^{2} - 4 \cdot 1 \cdot 0 = 1

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 0, x = - 1

- t > 0

\frac{8}{- 4}

\frac{5 !}{2 ! ( 5 - 2 ) !}

\frac{25}{49}

(4^{- 2})^{5}

What are the solutions to the equation (x+7}{12}-\frac{x^2+1)/4 =1-(x^2+2)/3 ?
The solutions to the equation (x+7}{12}-\frac{x^2+1)/4 =1-(x^2+2)/3 are x=0,x=-1
Find the zeros of (x+7}{12}-\frac{x^2+1)/4 =1-(x^2+2)/3
The zeros of (x+7}{12}-\frac{x^2+1)/4 =1-(x^2+2)/3 are x=0,x=-1