Solution to Question 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 Question 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 + √(13) ] / 2 and x2 = [ 1 - √(13) ] / 2

Solution to Question 3

• The given equation
  -(x + 2)(x - 1)=3
  
• Expand the left hand 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 Question 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 + √(13) ] / 2 and x2 = [ 1 - √(13) ] / 2

Solution to Question 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 = √7 and x2 = -√7

Solution to Question 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 = √2 and x - 2 = -√2
  
• Solve for x both equations.
  x1 = 2 + √2 and x2 = 2 - √2

Solution to Question 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 Question 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 - √(23) ] / 2 and x = [ 9 + √(23) ]

Solution to Question 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 Question 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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