Solve \frac{x^2-4}{x+2}+\frac{x-3}{x+2}

Question

Simplify the expression

\frac{x ^{2} - 7 + x}{x + 2}

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x = - 2

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x_{1} = - \frac{1 + 29}{2}, x_{2} = \frac{- 1 + 29}{2}

Alternative Form

x_{1} \approx - 3.192582, x_{2} \approx 2.192582

Evaluate

\frac{x ^{2} - 4}{x + 2} + \frac{x - 3}{x + 2}

To find the roots of the expression,set the expression equal to 0

\frac{x ^{2} - 4}{x + 2} + \frac{x - 3}{x + 2} = 0

Find the domain

More Steps

\frac{x ^{2} - 4}{x + 2} + \frac{x - 3}{x + 2} = 0, x \neq = - 2

Calculate

\frac{x ^{2} - 4}{x + 2} + \frac{x - 3}{x + 2} = 0

Simplify

More Steps

x - 2 + \frac{x - 3}{x + 2} = 0

Calculate the sum or difference

More Steps

\frac{x ^{2} + x - 7}{x + 2} = 0

Cross multiply

x^{2} + x - 7 = (x + 2) \times 0

Simplify the equation

x^{2} + x - 7 = 0

Substitute a= 1,b= 1 and c= - 7 into the quadratic formula x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

x = \frac{- 1 \pm 1 ^{2} - 4 ( - 7 )}{2}

Simplify the expression

More Steps

x = \frac{- 1 \pm 29}{2}

Separate the equation into 2 possible cases

x = \frac{- 1 + 29}{2} x = \frac{- 1 - 29}{2}

Use \frac{- a}{b} = \frac{a}{- b} = - \frac{a}{b} to rewrite the fraction

x = \frac{- 1 + 29}{2} x = - \frac{1 + 29}{2}

Check if the solution is in the defined range

x = \frac{- 1 + 29}{2} x = - \frac{1 + 29}{2}, x \neq = - 2

Find the intersection of the solution and the defined range

x = \frac{- 1 + 29}{2} x = - \frac{1 + 29}{2}

Solution

x_{1} = - \frac{1 + 29}{2}, x_{2} = \frac{- 1 + 29}{2}

Alternative Form

x_{1} \approx - 3.192582, x_{2} \approx 2.192582

Show Solution