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chels1226
-2 x+4 ----- = ------ x+2 x^2-4
A. x=0, -2 B. x=0,2,-2 C.x=2,-2 D.x=0
Solve for x over the real numbers: -2/(x+2) = (x+4)/(x^2-4) Multiply both sides by a polynomial to clear fractions. Cross multiply: -2 (x^2-4) = (x+2) (x+4) Write the quadratic polynomial on the left hand side in standard form. Expand out terms of the left hand side: 8-2 x^2 = (x+2) (x+4) Write the quadratic polynomial on the right hand side in standard form. Expand out terms of the right hand side: 8-2 x^2 = x^2+6 x+8 Move everything to the left hand side. Subtract x^2+6 x+8 from both sides: -6 x-3 x^2 = 0 Factor the left hand side. Factor x and constant terms from the left hand side: -3 x (x+2) = 0 Divide both sides by a constant to simplify the equation. Divide both sides by -3: x (x+2) = 0 Solve each term in the product separately. Split into two equations: x = 0 or x+2 = 0 Look at the second equation: Solve for x. Subtract 2 from both sides: x = 0 or x = -2 Now test that these solutions are correct by substituting into the original equation. Check the solution x = -2. -2/(x+2) => -2/(2-2) = infinity^~ ~~ infinity^~ (x+4)/(x^2-4) => (4-2)/(4-4) = infinity^~ ~~ infinity^~: So this solution is incorrect Check the solution x = 0. -2/(x+2) => -2/(2+0) = -1 (x+4)/(x^2-4) => (4+0)/(0-4) = -1: So this solution is correct Gather any correct solutions. The solution is: Answer: x = 0
(x+2)*(x+4)=(x^2-4)*(-2) x^2+4x+2x+8=-2x^2+8 3x^2+6x=0 delta=6^2-4*3*0=36 x1,2=[-6+-sqrt(delta)]/6 so x1=0 and x2=-2 so A is the correct answer