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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
.
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