Quadratic and Rational Equations: Step-by-Step Solutions

Practice Questions with Solutions

Question 1

Find all real solutions to the quadratic equation:

\[x^2 + 2x = -1\]

View Solution

• Rewrite with zero on the right: \[ x^2 + 2x + 1 = 0 \]
• Factor the quadratic: \[ (x + 1)^2 = 0 \]
• Solution (repeated root): \[ x = -1 \]

Question 2

Find all real solutions to the quadratic equation:

\[x^2 + 2 = x + 5\]

View Solution

• Rewrite with zero on the right: \[ x^2 - x - 3 = 0 \]
• Discriminant: \[ D = b^2 - 4ac = (-1)^2 - 4(1)(-3) = 13 \]
• Solutions: \[ x_1 = \frac{1 + \sqrt{13}}{2}, \quad x_2 = \frac{1 - \sqrt{13}}{2} \]

Question 3

Find all real solutions to the equation:

\[-(x + 2)(x - 1) = 3\]

View Solution

• 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.

Question 4

Find all real solutions to the rational equation:

\[\frac{2x + 1}{x + 2} = x - 1\]

View Solution

• 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: \[ x_1 = \frac{1 + \sqrt{13}}{2}, \quad x_2 = \frac{1 - \sqrt{13}}{2} \]

Question 5

Find all real solutions to the rational equation:

\[\frac{2}{x + 1} - \frac{1}{x - 2} = -1\]

View Solution

• 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: \[ x_1 = \sqrt{7}, \quad x_2 = -\sqrt{7} \]

Question 6

Find all real solutions to the quadratic equation:

\[2(x - 2)^2 - 6 = -2\]

View Solution

• 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: \[ x_1 = 2 + \sqrt{2}, \quad x_2 = 2 - \sqrt{2} \]

Question 7

Find all real solutions to the rational equation:

\[\frac{x}{x + 4} = -\frac{3}{x - 2} + \frac{18}{(x - 2)(x + 4)}\]

View Solution

• 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: \[ x = -3 \]

Question 8

Find all real solutions to the quadratic equation:

\[x^2 - 3(x - 3)^2 = 2\]

View Solution

• Expand: \[ x^2 - 3(x^2 - 6x + 9) = 2 \quad \Rightarrow \quad -2x^2 + 18x - 27 = 2 \]
• Rewrite: \[ -2x^2 + 18x - 29 = 0 \]
• Multiply by -1 for a positive leading coefficient: \[ 2x^2 - 18x + 29 = 0 \]
• Discriminant: \[ D = (-18)^2 - 4(2)(29) = 324 - 232 = 92 \]
• Solutions: \[ x = \frac{18 \pm \sqrt{92}}{4} = \frac{18 \pm 2\sqrt{23}}{4} = \frac{9 \pm \sqrt{23}}{2} \]

Question 9

Find all real solutions to the rational equation:

\[\frac{1}{x - 4} + \frac{1}{x + 4} = \frac{x^2}{x^2 - 16}\]

View Solution

• LCM: \(x^2 - 16 = (x - 4)(x + 4)\). Multiply both sides by \( (x - 4)(x + 4) \) : \[ (x + 4) + (x - 4) = x^2 \quad \Rightarrow \quad 2x = x^2 \]
• Rewrite: \[ x^2 - 2x = 0 \quad \Rightarrow \quad x(x - 2) = 0 \]
• Solutions: \[ x = 0, \quad x = 2 \]

Question 10

Find all real solutions to the rational equation:

\[-\frac{x}{x + 3} - \frac{x}{x - 3} = -\frac{4}{x^2 - 9} - \frac{1}{x + 3}\]

View Solution

• LCM: \(x^2 - 9 = (x + 3)(x - 3)\). Multiply both sides: \[ -x(x - 3) - x(x + 3) = -4 - (x - 3) \]
• Simplify and rewrite: \[ -x^2 + 3x - x^2 - 3x = -4 - x + 3 \quad \Rightarrow \quad -2x^2 = -x - 1 \quad \Rightarrow \quad 2x^2 - x - 1 = 0 \]
• Factor and solve: \[ (2x + 1)(x - 1) = 0 \]
• Solutions: \[ x = 1, \quad x = -\frac{1}{2} \]