Simple Quadratic Equation

Solve $x^2 + 6x + 5 = 0$ using the quadratic formula.

Identify a, b, and c: Comparing to $ax^2 + bx + c = 0$: $a = 1$, $b = 6$, $c = 5$ = $a=1, b=6, c=5$

Calculate the discriminant: $b^2 - 4ac = 6^2 - 4(1)(5) = 36 - 20 = 16$ = $D = 16$

Apply the formula: $x = \frac{-6 \pm \sqrt{16}}{2(1)} = \frac{-6 \pm 4}{2}$ = $x = \frac{-6 \pm 4}{2}$

Find both solutions: $x_1 = \frac{-6 + 4}{2} = \frac{-2}{2} = -1$ $x_2 = \frac{-6 - 4}{2} = \frac{-10}{2} = -5$ = $x = -1$ or $x = -5$

Answer: $x = -1$ or $x = -5$

Equation with Non-Unit Leading Coefficient

Solve $2x^2 - 7x + 3 = 0$ using the quadratic formula.

Identify a, b, and c: $a = 2$, $b = -7$, $c = 3$ Note: $b$ is negative! = $a=2, b=-7, c=3$

Calculate the discriminant: $b^2 - 4ac = (-7)^2 - 4(2)(3) = 49 - 24 = 25$ = $D = 25$

Apply the formula: $x = \frac{-(-7) \pm \sqrt{25}}{2(2)} = \frac{7 \pm 5}{4}$ = $x = \frac{7 \pm 5}{4}$

Find both solutions: $x_1 = \frac{7 + 5}{4} = \frac{12}{4} = 3$ $x_2 = \frac{7 - 5}{4} = \frac{2}{4} = \frac{1}{2}$ = $x = 3$ or $x = \frac{1}{2}$

Answer: $x = 3$ or $x = \frac{1}{2}$

Equation with Irrational Solutions

Solve $x^2 - 4x - 1 = 0$ using the quadratic formula.

Identify a, b, and c: $a = 1$, $b = -4$, $c = -1$ = $a=1, b=-4, c=-1$

Calculate the discriminant: $b^2 - 4ac = (-4)^2 - 4(1)(-1) = 16 + 4 = 20$ = $D = 20$

Apply the formula: $x = \frac{-(-4) \pm \sqrt{20}}{2(1)} = \frac{4 \pm \sqrt{20}}{2}$ = $x = \frac{4 \pm \sqrt{20}}{2}$

Simplify the radical: $\sqrt{20} = \sqrt{4 \cdot 5} = 2\sqrt{5}$ $x = \frac{4 \pm 2\sqrt{5}}{2} = 2 \pm \sqrt{5}$ = $x = 2 + \sqrt{5}$ or $x = 2 - \sqrt{5}$

Answer: $x = 2 + \sqrt{5} \approx 4.24$ or $x = 2 - \sqrt{5} \approx -0.24$

Understanding the Discriminant

How many solutions does $x^2 + 2x + 5 = 0$ have?

Identify a, b, and c: $a = 1$, $b = 2$, $c = 5$ = $a=1, b=2, c=5$

Calculate the discriminant: $b^2 - 4ac = 2^2 - 4(1)(5) = 4 - 20 = -16$ = $D = -16$

Interpret the discriminant: Since $D < 0$, we would need $\sqrt{-16}$, which is not a real number = No real solutions

Conclusion: When the discriminant is negative, the parabola doesn't cross the x-axis = Zero real solutions

Answer: This equation has no real solutions because the discriminant is negative ($D = -16$).

Mistake: Forgetting to make b negative when it's already negative

Why: If $b = -7$, then $-b = -(-7) = 7$, not $-7$. The formula says $-b$, so you negate whatever b is.

Correct: Always write $-b$ first, then substitute: if $b = -7$, write $-(-7) = 7$

Mistake: Calculating $b^2$ incorrectly when b is negative

Why: Students sometimes write $-7^2 = -49$ instead of $(-7)^2 = 49$

Correct: The entire coefficient is squared: $(-7)^2 = (-7) \times (-7) = 49$

Mistake: Dividing only part of the numerator by 2a

Why: Writing $\frac{-b}{2a} \pm \sqrt{b^2-4ac}$ instead of $\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

Correct: The entire numerator $(-b \pm \sqrt{...})$ is divided by $2a$

Mistake: Stopping at one solution

Why: The $\pm$ means plus OR minus, giving two potential solutions

Correct: Always calculate both: $x_1 = \frac{-b + \sqrt{D}}{2a}$ and $x_2 = \frac{-b - \sqrt{D}}{2a}$

Projectile Motion

When you throw a ball, its height follows a quadratic equation. The formula helps find when it lands.

A ball thrown upward has height $h = -5t^2 + 20t + 1$ meters after $t$ seconds. To find when it hits the ground ($h = 0$), solve $-5t^2 + 20t + 1 = 0$ using the formula.

Business Profit Optimization

Companies use quadratic equations to model profit and find break-even points.

A company's profit is $P = -2x^2 + 100x - 800$ dollars, where $x$ is units sold. To find break-even points (where $P = 0$), use the quadratic formula.

Q: When should I use the quadratic formula instead of factoring?

A: Use the quadratic formula when: (1) the equation doesn't factor easily, (2) the coefficients are large or decimals, or (3) you want a reliable method that always works. Factoring is faster when it works, but the formula is more universal.

Q: What does it mean when the discriminant is zero?

A: When $b^2 - 4ac = 0$, the equation has exactly one solution (called a repeated or double root). Graphically, the parabola just touches the x-axis at one point.

Q: Why is there a plus-minus sign in the formula?

A: The $\pm$ comes from taking the square root. Since both $5^2 = 25$ and $(-5)^2 = 25$, the square root of 25 could be either +5 or -5. This gives us two possible values for x.