The solutions to quadratic and rational equations are presented.
Solution to Problem 1: The given equation x2 + 2x = -1 Write the above quadratic equation with right side equal to 0. x2 + 2x + 1 = 0 Factor the equation. (x + 1)2 = 0 Solve the equation to obtain the repeated solution. x = -1 Solution to Problem 2: The given equation is a quadratic one x2 + 2 = x + 5 Write the equation with right side equal to 0. x2 - x - 3 = 0 Find the discriminant D of the quadratic equation. D = b 2 - 4 ac = 13 Solve the above equation to obtain 2 real solutions. x1 = [ 1 + sqrt(13) ] / 2 and x2 = [ 1 - sqrt(13) ] / 2 Solution to Problem 3: The given equation -(x + 2)(x - 1)=3 Expand the right term of the equation. -x2 - x + 2 = 3 Rewrite the above equation with right side equal to 0. -x2 - x - 1 = 0 Find discriminant D. D = 1 - 4(-1)(-1) = -3 Since we are asked to find real solutions and the discriminant is negative, this equation has no real solutions. Solution to Problem 4: The equation to solve is (2x + 1) / (x + 2) = x - 1 The domain of the rational expression on the left of the equal sign of the equation is all real numbers except -2. Multiply both sides of the equation by (x + 2) and simplify. (2x + 1) = (x - 1)(x + 2) Expand the right side, group like terms and write the equation in standard form. x2 - x - 3 = 0 Find discriminant D. D = 1 - 4(1)(-3) = 13 and the solutions are. x1 = [ 1 + sqrt(13) ] / 2 and x2 = [ 1 - sqrt(13) ] / 2 Solution to Problem 5: The equation to solve is 2 / (x + 1) - 1 / (x - 2) = -1 The LCM of the denominators of the rational expressions is. lcm = (x + 1)(x - 2) We now multiply both sides of the equations by the lcm and simplify. 2(x - 2) - 1(x + 1) = -1(x + 1)(x - 2) Expand the right side and group. x - 5 = -x2 + x + 2 Write with right side equal to 0. x2 = 7 The solutions are. x1 = sqrt(7) and x2 = -sqrt(7) Solution to Problem 6: The given equation is 2(x - 2)2 - 6 = -2 Add 6 to both sides and simplify. 2(x - 2)2 = 4 Divide both side by 2. (x - 2)2 = 2 Extract the square root to obtain. x - 2 = sqrt(2) and x - 2 = -sqrt(2) Solve for x both equations. x1 = 2 + sqrt(2) and x2 = 2 - sqrt(2) Solution to Problem 7: The given equation is x / (x + 4) = -3 / (x - 2) + 18 / (x - 2) (x + 4) The lcm of the denominators of the rationa expression is equal to lcm = (x + 4)(x - 2) Multiply all terms by the lcm and simplify. x (x - 2) = -3 (x + 4) + 18 Expand and group. x 2 - 2x = -3x -12 + 18 Write the equation with right side equal to 0. x 2 + x - 6 = 0 Factor and solve. (x + 3)(x - 2) = 0 The solutions to the last equation are x = -3 and x = 2. The solution x = 2 cannot be a solution to the given equation as it makes the denominator equal to 0. So the only solution to the given equation is x = -3. Solution to Problem 8: The equation to solve is x2 - 3(x - 3)2 = 2 Expand the term 3(x - 3)2 and group x2 - 3(x2 -6x + 9) = 2 -2x2 + 18x - 27 = 2 Write the equation with right side equal to 0. -2x2 + 18x - 29 = 0 The discriminant D of the above quadratic equation is equal to . D = 92 The solutions to the given equation are x = [ 9 - sqrt(23) ] / 2 and x = [ 9 + sqrt(23) ] Solution to Problem 9: The equation to solve is 1 / (x - 4) + 1 / (x + 4)= x2 / (x2 - 16) The lcm of the denominators of the rational expressions is given by lcm = (x - 4)(x + 4) = x2 - 16 Multiply all terms by the lcm and simplify. x + 4 + x - 4 = x2 Simplify and write equation with right term equal to 0. x2 - 2 x = 0 Factor and solve x(x - 2) = 0 x = 0 and x = 2 are the solutions. Solution to Problem 10: The equation to solve is -x / (x + 3) - x / (x - 3) = - 4 / (x2 - 9) - 1 / (x + 3) The lcm of the denominators of the rational expressions is given by lcm = (x - 3)(x + 3) = x2 - 9 Multiply all terms of the equation by the lcm and simplify. -x(x - 3) - x(x + 3) = -4 -(x - 3) Expand, group like terms and with right term equal to 0. 2x2 - x - 1 = 0 Solve the above quadratic equation to obtain the solutions x = 1 and x = -1/2 More references and links to tutorials and problems on equations. Solve Quadratic Equations Using Discriminants. Tutorial on Equations of the Quadratic Form. More Math Problems, Questions and Online Self Tests.
Solution to Problem 1: The given equation x2 + 2x = -1 Write the above quadratic equation with right side equal to 0. x2 + 2x + 1 = 0 Factor the equation. (x + 1)2 = 0 Solve the equation to obtain the repeated solution. x = -1
Solution to Problem 2: The given equation is a quadratic one x2 + 2 = x + 5 Write the equation with right side equal to 0. x2 - x - 3 = 0 Find the discriminant D of the quadratic equation. D = b 2 - 4 ac = 13 Solve the above equation to obtain 2 real solutions. x1 = [ 1 + sqrt(13) ] / 2 and x2 = [ 1 - sqrt(13) ] / 2
Solution to Problem 3: The given equation -(x + 2)(x - 1)=3 Expand the right term of the equation. -x2 - x + 2 = 3 Rewrite the above equation with right side equal to 0. -x2 - x - 1 = 0 Find discriminant D. D = 1 - 4(-1)(-1) = -3 Since we are asked to find real solutions and the discriminant is negative, this equation has no real solutions.
Solution to Problem 4: The equation to solve is (2x + 1) / (x + 2) = x - 1 The domain of the rational expression on the left of the equal sign of the equation is all real numbers except -2. Multiply both sides of the equation by (x + 2) and simplify. (2x + 1) = (x - 1)(x + 2) Expand the right side, group like terms and write the equation in standard form. x2 - x - 3 = 0 Find discriminant D. D = 1 - 4(1)(-3) = 13 and the solutions are. x1 = [ 1 + sqrt(13) ] / 2 and x2 = [ 1 - sqrt(13) ] / 2
Solution to Problem 5: The equation to solve is 2 / (x + 1) - 1 / (x - 2) = -1 The LCM of the denominators of the rational expressions is. lcm = (x + 1)(x - 2) We now multiply both sides of the equations by the lcm and simplify. 2(x - 2) - 1(x + 1) = -1(x + 1)(x - 2) Expand the right side and group. x - 5 = -x2 + x + 2 Write with right side equal to 0. x2 = 7 The solutions are. x1 = sqrt(7) and x2 = -sqrt(7)
Solution to Problem 6: The given equation is 2(x - 2)2 - 6 = -2 Add 6 to both sides and simplify. 2(x - 2)2 = 4 Divide both side by 2. (x - 2)2 = 2 Extract the square root to obtain. x - 2 = sqrt(2) and x - 2 = -sqrt(2) Solve for x both equations. x1 = 2 + sqrt(2) and x2 = 2 - sqrt(2)
Solution to Problem 7: The given equation is x / (x + 4) = -3 / (x - 2) + 18 / (x - 2) (x + 4) The lcm of the denominators of the rationa expression is equal to lcm = (x + 4)(x - 2) Multiply all terms by the lcm and simplify. x (x - 2) = -3 (x + 4) + 18 Expand and group. x 2 - 2x = -3x -12 + 18 Write the equation with right side equal to 0. x 2 + x - 6 = 0 Factor and solve. (x + 3)(x - 2) = 0 The solutions to the last equation are x = -3 and x = 2. The solution x = 2 cannot be a solution to the given equation as it makes the denominator equal to 0. So the only solution to the given equation is x = -3.
Solution to Problem 8: The equation to solve is x2 - 3(x - 3)2 = 2 Expand the term 3(x - 3)2 and group x2 - 3(x2 -6x + 9) = 2 -2x2 + 18x - 27 = 2 Write the equation with right side equal to 0. -2x2 + 18x - 29 = 0 The discriminant D of the above quadratic equation is equal to . D = 92 The solutions to the given equation are x = [ 9 - sqrt(23) ] / 2 and x = [ 9 + sqrt(23) ]
Solution to Problem 9: The equation to solve is 1 / (x - 4) + 1 / (x + 4)= x2 / (x2 - 16) The lcm of the denominators of the rational expressions is given by lcm = (x - 4)(x + 4) = x2 - 16 Multiply all terms by the lcm and simplify. x + 4 + x - 4 = x2 Simplify and write equation with right term equal to 0. x2 - 2 x = 0 Factor and solve x(x - 2) = 0 x = 0 and x = 2 are the solutions.
Solution to Problem 10: The equation to solve is -x / (x + 3) - x / (x - 3) = - 4 / (x2 - 9) - 1 / (x + 3) The lcm of the denominators of the rational expressions is given by lcm = (x - 3)(x + 3) = x2 - 9 Multiply all terms of the equation by the lcm and simplify. -x(x - 3) - x(x + 3) = -4 -(x - 3) Expand, group like terms and with right term equal to 0. 2x2 - x - 1 = 0 Solve the above quadratic equation to obtain the solutions x = 1 and x = -1/2
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