Solutions to Quadratic Equations With Rational Expressions
Free Mathematics Tutorials
Home
Solutions to Quadratic Equations With Rational Expressions
The solutions to questions on
quadratic and rational equations
are presented.
Solution to Question 1
The given equation
x
^{2}
+ 2x = -1
Write the above quadratic equation with right side equal to 0.
x
^{2}
+ 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
x
^{2}
+ 2 = x + 5
Write the equation with right side equal to 0.
x
^{2}
- 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.
-x
^{2}
- x + 2 = 3
Rewrite the above equation with right side equal to 0.
-x
^{2}
- 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.
x
^{2}
- 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 = -x
^{2}
+ x + 2
Write with right side equal to 0.
x
^{2}
= 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
x
^{2}
- 3(x - 3)
^{2}
= 2
Expand the term 3(x - 3)
^{2}
and group
x
^{2}
- 3(x
^{2}
-6x + 9) = 2
-2x
^{2}
+ 18x - 27 = 2
Write the equation with right side equal to 0.
-2x
^{2}
+ 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)= x
^{2}
/ (x
^{2}
- 16)
The lcm of the denominators of the rational expressions is given by
lcm = (x - 4)(x + 4) = x
^{2}
- 16
Multiply all terms by the lcm and simplify.
x + 4 + x - 4 = x
^{2}
Simplify and write equation with right term equal to 0.
x
^{2}
- 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 / (x
^{2}
- 9) - 1 / (x + 3)
The lcm of the denominators of the rational expressions is given by
lcm = (x - 3)(x + 3) = x
^{2}
- 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.
2x
^{2}
- 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 questions on equations
Solve Quadratic Equations Using Discriminants
Tutorial on Equations of the Quadratic Form.
Math Problems, Questions and Online Self Tests
.
facebook
twitter
Search
Popular Pages
Home