Solve the Quadratic Equation: x² - 3x - 18 = 0

Q: How do I know which two numbers multiply to -18 and add to -3?

List all factor pairs of -18: (1,-18), (-1,18), (2,-9), (-2,9), (3,-6), (-3,6). Check which pair adds to -3: only (-6) + 3 = -3, so use -6 and 3!

Q: Why does (x-6)(x+3) = 0 give me x = 6 and x = -3?

When a product equals zero, at least one factor must be zero. So either x-6=0 (giving x=6) or x+3=0 (giving x=-3). Both solutions are valid!

Q: What if I can't factor the quadratic easily?

Try the quadratic formula: $ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} $. For $ x^2-3x-18=0 $, you'd get the same answers: x = 6 and x = -3.

Q: How can I check if my factoring is correct?

Expand your factors using FOIL: $ (x-6)(x+3) = x^2 + 3x - 6x - 18 = x^2 - 3x - 18 $. If you get back the original equation, your factoring is right!

Q: Do I always get exactly two solutions for quadratic equations?

Usually yes! Most quadratics have two distinct solutions. Sometimes you might get one repeated solution (like x = 2, x = 2), and rarely no real solutions at all.

Quadratic Equations with Factoring Method

$x^2-3x-18=0$

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:05 Let's find X together.

00:08 We'll break it down using a trinomial approach. First, let's identify the coefficients.

00:14 We're looking for two numbers. Their sum should equal B, which is negative three.

00:19 Next, their product needs to equal C, which is negative eighteen.

00:25 Once we find these numbers, we'll substitute them into parentheses.

00:34 Now, let's find what makes each factor equal to zero.

00:45 And that's how we solve this problem!

Step-by-step written solution

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Understand the problem

$x^2-3x-18=0$

Step-by-step solution

Let's observe that the given equation:

$x^2-3x-18=0$ is a quadratic equation that can be solved using quick factoring:

$x^2-3x-18=0 \longleftrightarrow\begin{cases}\underline{?}\cdot\underline{?}=-18\\ \underline{?}+\underline{?}=-3\end{cases}\\ \downarrow\\ (x-6)(x+3)=0$ and therefore we get two simpler equations from which we can extract the solution:

$(x-6)(x+3)=0 \\ \downarrow\\ x-6=0\rightarrow\boxed{x=6}\\ x+3=0\rightarrow\boxed{x=-3}\\ \boxed{x=6,-3}$ Therefore, the correct answer is answer A.

Final Answer

$x=-3,x=6$

Key Points to Remember

Essential concepts to master this topic

Rule: Find two numbers that multiply to constant and add to coefficient
Technique: For $x^2-3x-18$ , need factors of -18 that add to -3
Check: Substitute solutions: $6^2-3(6)-18 = 0$ and $(-3)^2-3(-3)-18 = 0$ ✓

Common Mistakes

Avoid these frequent errors

Finding factors that multiply to the coefficient instead of the constant
Don't look for factors of -3 when the constant is -18 = completely wrong factors! This gives impossible solutions that don't work. Always find two numbers that multiply to the constant term (-18) and add to the linear coefficient (-3).

Practice Quiz

Test your knowledge with interactive questions

Solve the following expression:

$$ x^2-1=0 $$

FAQ

Everything you need to know about this question

How do I know which two numbers multiply to -18 and add to -3?

List all factor pairs of -18: (1,-18), (-1,18), (2,-9), (-2,9), (3,-6), (-3,6). Check which pair adds to -3: only (-6) + 3 = -3, so use -6 and 3!

Why does (x-6)(x+3) = 0 give me x = 6 and x = -3?

When a product equals zero, at least one factor must be zero. So either x-6=0 (giving x=6) or x+3=0 (giving x=-3). Both solutions are valid!

What if I can't factor the quadratic easily?

Try the quadratic formula: $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ . For $x^2-3x-18=0$ , you'd get the same answers: x = 6 and x = -3.

How can I check if my factoring is correct?

Expand your factors using FOIL: $(x-6)(x+3) = x^2 + 3x - 6x - 18 = x^2 - 3x - 18$ . If you get back the original equation, your factoring is right!

Do I always get exactly two solutions for quadratic equations?

Usually yes! Most quadratics have two distinct solutions. Sometimes you might get one repeated solution (like x = 2, x = 2), and rarely no real solutions at all.

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