Q. \(x^2-3x-18\)
Answer
Factor the quadratic.
\[
x^2-3x-18=(x-6)(x+3)
\]
Set each factor equal to zero:
\[
x-6=0 \Rightarrow x=6,\quad x+3=0 \Rightarrow x=-3
\]
Final result: \(x=6\) or \(x=-3\).
Detailed Explanation
We want to solve the expression \(x^2 – 3x – 18\). A common first step is to factor it, if possible.
Step 1: Factor the quadratic.
For a quadratic of the form \(ax^2 + bx + c\), we look for two numbers that:
- Multiply to \(ac\), which is \((-18)\)
- Add to \(b\), which is \((-3)\)
Here, \(a = 1\), \(b = -3\), and \(c = -18\). So we need two numbers that multiply to \((1)(-18) = -18\) and add to \(-3\).
Step 2: Find the numbers.
List factor pairs of \(-18\):
- \(1\) and \(-18\) (sum \(-17\))
- \(2\) and \(-9\) (sum \(-7\))
- \(3\) and \(-6\) (sum \(-3\))
The pair \(3\) and \(-6\) works because:
- \(3 \cdot (-6) = -18\)
- \(3 + (-6) = -3\)
Step 3: Write the factored form.
Using these values, we can factor:
\[
x^2 – 3x – 18 = (x + 3)(x – 6)
\]
Final Answer:
\[
x^2 – 3x – 18 = (x + 3)(x – 6)
\]
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Algebra FAQ
Factor \(x^2-3x-18\).
\((x-6)(x+3)\).
Solve \(x^2-3x-18=0\).
Set \((x-6)(x+3)=0\), so \(x=6\) or \(x=-3\).
What are the roots of \(x^2-3x-18\)?
The roots are \(\{6,-3\}\).
Find the vertex of \(y=x^2-3x-18\).
\(x_v=-\frac{b}{2a}=-\frac{-3}{2}=\frac{3}{2}\). Then \(y_v=f\!\left(\frac{3}{2}\right)=-\frac{45}{4}\).
What is the discriminant of \(x^2-3x-18\)?
\(a=1,b=-3,c=-18\). So \(D=b^2-4ac=9-4(1)(-18)=81\).
Compute \(f(3)\) and \(f(-3)\) for \(f(x)=x^2-3x-18\).
\(f(3)=9-9-18=-18\). \(f(-3)=9+9-18=0\).
Use the tools to factor x²-3x-18.
Check your roots and simplify.
Check your roots and simplify.
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