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.