Solutions to Quadratic Equations and Rational Expressions
This page presents step-by-step solutions to quadratic and rational equations. Each solution includes explanations to help understand the process.
Solution to Question 1
- Given equation:
\[
x^2 + 2x = -1
\]
- Rewrite with zero on the right:
\[
x^2 + 2x + 1 = 0
\]
- Factor the quadratic:
\[
(x + 1)^2 = 0
\]
- Solution (repeated root):
\[
x = -1
\]
Solution to Question 2
- Given equation:
\[
x^2 + 2 = x + 5
\]
- Rewrite with zero on the right:
\[
x^2 - x - 3 = 0
\]
- Discriminant:
\[
D = b^2 - 4ac = (-1)^2 - 4(1)(-3) = 13
\]
- Solutions:
\[
x_1 = \frac{1 + \sqrt{13}}{2}, \quad x_2 = \frac{1 - \sqrt{13}}{2}
\]
Solution to Question 3
- Given equation:
\[
-(x + 2)(x - 1) = 3
\]
- Expand:
\[
-x^2 - x + 2 = 3
\]
- Rewrite with zero on the right:
\[
-x^2 - x - 1 = 0 \quad \Rightarrow \quad x^2 + x + 1 = 0
\]
- Discriminant:
\[
D = 1^2 - 4(1)(1) = -3
\]
- Since \(D < 0\), there are no real solutions.
Solution to Question 4
- Given equation:
\[
\frac{2x + 1}{x + 2} = x - 1
\]
- Domain restriction: \(x \neq -2\). Multiply both sides by \(x + 2\):
\[
2x + 1 = (x - 1)(x + 2)
\]
- Expand and simplify:
\[
2x + 1 = x^2 + x - 2 \quad \Rightarrow \quad x^2 - x - 3 = 0
\]
- Discriminant:
\[
D = (-1)^2 - 4(1)(-3) = 13
\]
- Solutions:
\[
x_1 = \frac{1 + \sqrt{13}}{2}, \quad x_2 = \frac{1 - \sqrt{13}}{2}
\]
Solution to Question 5
- Given equation:
\[
\frac{2}{x + 1} - \frac{1}{x - 2} = -1
\]
- LCM of denominators: \((x + 1)(x - 2)\). Multiply both sides:
\[
2(x - 2) - (x + 1) = -1 (x + 1)(x - 2)
\]
- Expand and simplify:
\[
x - 5 = -x^2 + x + 2 \quad \Rightarrow \quad x^2 = 7
\]
- Solutions:
\[
x_1 = \sqrt{7}, \quad x_2 = -\sqrt{7}
\]
Solution to Question 6
- Given equation:
\[
2(x - 2)^2 - 6 = -2
\]
- Add 6:
\[
2(x - 2)^2 = 4
\]
- Divide by 2:
\[
(x - 2)^2 = 2
\]
- Take square roots:
\[
x - 2 = \sqrt{2} \quad \text{or} \quad x - 2 = -\sqrt{2}
\]
- Solutions:
\[
x_1 = 2 + \sqrt{2}, \quad x_2 = 2 - \sqrt{2}
\]
Solution to Question 7
- Given equation:
\[
\frac{x}{x + 4} = \frac{-3}{x - 2} + \frac{18}{(x - 2)(x + 4)}
\]
- LCM of denominators: \((x + 4)(x - 2)\). Multiply both sides:
\[
x(x - 2) = -3(x + 4) + 18
\]
- Simplify:
\[
x^2 - 2x = -3x - 12 + 18 \quad \Rightarrow \quad x^2 + x - 6 = 0
\]
- Factor and solve:
\[
(x + 3)(x - 2) = 0 \quad \Rightarrow \quad x = -3, \, x = 2
\]
- Check restrictions: \(x = 2\) is excluded (denominator zero). So solution:
\[
x = -3
\]
Solution to Question 8
- Given equation:
\[
x^2 - 3(x - 3)^2 = 2
\]
- Expand:
\[
x^2 - 3(x^2 - 6x + 9) = 2 \quad \Rightarrow \quad -2x^2 + 18x - 27 = 2
\]
- Rewrite:
\[
-2x^2 + 18x - 29 = 0
\]
- Discriminant:
\[
D = 18^2 - 4(-2)(-29) = 324 - 232 = 92
\]
- Solutions:
\[
x = \frac{18 \pm \sqrt{92}}{-4} = \frac{9 \pm \sqrt{23}}{2}
\]
Solution to Question 9
- Given equation:
\[
\frac{1}{x - 4} + \frac{1}{x + 4} = \frac{x^2}{x^2 - 16}
\]
- LCM: \(x^2 - 16 = (x - 4)(x + 4)\). Multiply both sides:
\[
(x + 4) + (x - 4) = x^2 \quad \Rightarrow \quad 2x = x^2
\]
- Rewrite:
\[
x^2 - 2x = 0 \quad \Rightarrow \quad x(x - 2) = 0
\]
- Solutions:
\[
x = 0, \quad x = 2
\]
Solution to Question 10
- Given equation:
\[
-\frac{x}{x + 3} - \frac{x}{x - 3} = - \frac{4}{x^2 - 9} - \frac{1}{x + 3}
\]
- LCM: \(x^2 - 9 = (x + 3)(x - 3)\). Multiply both sides:
\[
-x(x - 3) - x(x + 3) = -4 - (x - 3)
\]
- Simplify and rewrite:
\[
2x^2 - x - 1 = 0
\]
- Solutions:
\[
x = 1, \quad x = -\frac{1}{2}
\]
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