Quadratic Equations – Intro and Solved Exercises
What Is a Quadratic Equation?
A quadratic equation is an equation where the variable appears with exponent 2:
\[ax^2 + bx + c = 0 \qquad \text{with } a \neq 0\]It has at most two real solutions.
Common Forms
Some quadratic equations look simpler than others:
- Standard form: ( ax^2 + bx + c = 0 )
- Pure quadratic: ( ax^2 + c = 0 )
- Spurious quadratic: ( ax^2 + bx = 0 )
- Perfect square: ( (x + r)^2 = 0 )
Solving Methods
✅ Factoring and the Zero Product Property (ZPP)
If the equation can be factored as:
\[(x - r_1)(x - r_2) = 0\]Then the solutions are:
\[x = r_1 \quad \text{and} \quad x = r_2\]✅ Quadratic Formula
For any equation in standard form:
\[ax^2 + bx + c = 0\]You can use:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]The part under the square root is called the discriminant:
\[\Delta = b^2 - 4ac\]✅ How Many Solutions?
- If ( \Delta > 0 ): two real solutions
- If ( \Delta = 0 ): one real solution (a double root)
- If ( \Delta < 0 ): no real solution
Solved Examples
Example 1
Solve:
\[x^2 - 9x = 0\]Factor:
\[x(x - 9) = 0 \Rightarrow x = 0 \quad \text{or} \quad x = 9\]Example 2
Solve:
\[x^2 + 6x + 9 = 0\]This is a perfect square:
\[(x + 3)^2 = 0 \Rightarrow x = -3\]Example 3
Solve:
\[2x^2 + 3x + 1 = 0\]Use the formula:
\[x = \frac{-3 \pm \sqrt{3^2 - 4\cdot2\cdot1}}{2 \cdot 2} = \frac{-3 \pm \sqrt{1}}{4}\]So:
\[x = -1 \quad \text{or} \quad x = -\frac{1}{2}\]Practice Exercises
Try solving the following equations:
- ( x^2 - 4 = 0 )
- ( x(x - 5) = 0 )
- ( x^2 + 8x + 16 = 0 )
- ( x^2 + 2x - 3 = 0 )
- ( x^2 + x + 1 = 0 )
Suggested Solutions
- ( x = -2 ), ( x = 2 )
- ( x = 0 ), ( x = 5 )
- ( x = -4 )
- ( x = 1 ), ( x = -3 )
- ❌ No real solutions (discriminant is negative)
📚 Want the Full Lesson?
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👉 Quadratic Equations – Full Lesson (PDF)
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