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

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

List factor pairs of -18: (1,-18), (-1,18), (2,-9), (-2,9), (3,-6), (-3,6). Then check which pair adds to -3. Only 6 + (-9) = -3, but that's not right. Try (-3) + 6 = 3, no. Try 6 + (-3) = 3, no. Actually, 6 + (-9) = -3, so it's 6 and -9, but we need factors of -18!

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). This is called the zero product property.

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

If factoring seems difficult, you can always use the quadratic formula: $ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} $. But try factoring first - it's usually faster when it works!

Q: How do I check if my factored form is correct?

Expand your factored form using FOIL: $ (x-6)(x+3) = x^2 + 3x - 6x - 18 = x^2 - 3x - 18 $. If this matches your original equation, you factored correctly!

Q: Can a quadratic equation have more than two solutions?

No! A quadratic equation can have at most two real solutions. Sometimes it has two solutions, one solution (repeated), or no real solutions, but never more than two.

Quadratic Equations with Quick Factoring

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

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Find X

00:03 Let's break it down using trinomial, let's look at the coefficients

00:07 We want to find 2 numbers whose sum equals B (-3)

00:13 and their product equals C (-18)

00:20 These are the matching numbers, let's substitute in parentheses

00:30 Let's find what zeros each factor

00:39 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution

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 c and add to b
Technique: For $x^2-3x-18$ , find factors of -18 that add to -3: 6 and -3
Check: Substitute both solutions: $(6)^2-3(6)-18=0$ and $(-3)^2-3(-3)-18=0$ ✓

Common Mistakes

Avoid these frequent errors

Mixing up the sign patterns when factoring
Don't write (x-6)(x-3)=0 when you need factors that multiply to -18! This gives wrong solutions x=6,3 instead of x=6,-3. Always check: your factors must multiply to give the original constant term and add to give the coefficient of x.

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 factor pairs of -18: (1,-18), (-1,18), (2,-9), (-2,9), (3,-6), (-3,6). Then check which pair adds to -3. Only 6 + (-9) = -3, but that's not right. Try (-3) + 6 = 3, no. Try 6 + (-3) = 3, no. Actually, 6 + (-9) = -3, so it's 6 and -9, but we need factors of -18!

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). This is called the zero product property.

What if I can't factor the quadratic easily?

If factoring seems difficult, you can always use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ . But try factoring first - it's usually faster when it works!

How do I check if my factored form is correct?

Expand your factored form using FOIL: $(x-6)(x+3) = x^2 + 3x - 6x - 18 = x^2 - 3x - 18$ . If this matches your original equation, you factored correctly!

Can a quadratic equation have more than two solutions?

No! A quadratic equation can have at most two real solutions. Sometimes it has two solutions, one solution (repeated), or no real solutions, but never more than two.

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