Q. \(x^2 – 11x + 18\)

Q: Solve x^2-11x+18=0?

From factors (x-9)(x-2)=0, so x=9 or x=2.

Q: Find the roots using the quadratic formula?

x=\dfrac{11\pm\sqrt{121-72}}{2}=\dfrac{11\pm 7}{2}, giving x=9,2.

Q: Complete the square for x^2-11x+18?

x^2-11x=(x-\tfrac{11}{2})^2-\tfrac{121}{4}. Then x^2-11x+18=(x-\tfrac{11}{2})^2-\tfrac{49}{4}.

Q: What is the vertex of y=x^2-11x+18?

Vertex x-coordinate is h=\tfrac{11}{2}. Then y=\left(\tfrac{11}{2}\right)^2-11\cdot\tfrac{11}{2}+18=-\tfrac{49}{4}.

Q: What are the x-intercepts and y-intercept?

x-intercepts: x=2,9. y-intercept: set x=0, so y=18.

Answer

We factor the quadratic \(x^2 – 11x + 18\). Find two numbers that multiply to \(18\) and add to \(-11\): \(-2\) and \(-9\).

\[
x^2 – 11x + 18 = (x-2)(x-9)
\]

So the factored form is \((x-2)(x-9)\). The solutions are \(x=2\) and \(x=9\).

Detailed Explanation

We are asked to simplify or factor the expression \(x^2 – 11x + 18\). A standard method is to factor a quadratic into two binomials.

Step 1: Identify the quadratic form

We have

\[x^2 – 11x + 18.\]

This matches the form

\[x^2 + bx + c,\]

where \(b = -11\) and \(c = 18\).

Step 2: Find two numbers that multiply to \(18\)

We want two numbers whose product is \(18\). Also, their sum must be \(-11\) because the middle term is \(-11x\).

The factor pairs of \(18\) (with signs considered for the sum) are:

\[1 \cdot 18 = 18,\quad 2 \cdot 9 = 18,\quad 3 \cdot 6 = 18.\]

To get a sum of \(-11\), we try negative pairs:

\[-2 + (-9) = -11.\]

So the two numbers are \(-2\) and \(-9\).

Step 3: Write the factored form

Use these numbers as constants in the binomials:

\[x^2 – 11x + 18 = (x – 2)(x – 9).\]

Step 4: (Quick check by multiplying)

Multiply \((x – 2)(x – 9)\):

\[(x – 2)(x – 9) = x(x – 9) – 2(x – 9).\]

\[= x^2 – 9x – 2x + 18.\]

\[= x^2 – 11x + 18.\]

Final Answer

\[x^2 – 11x + 18 = (x – 2)(x – 9).\]

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Algebra FAQ

Factor \(x^2-11x+18\)?

\(x^2-11x+18=(x-9)(x-2)\).

Solve \(x^2-11x+18=0\)?

From factors \((x-9)(x-2)=0\), so \(x=9\) or \(x=2\).

Find the roots using the quadratic formula?

\(x=\dfrac{11\pm\sqrt{121-72}}{2}=\dfrac{11\pm 7}{2}\), giving \(x=9,2\).

Complete the square for \(x^2-11x+18\)?

\(x^2-11x=(x-\tfrac{11}{2})^2-\tfrac{121}{4}\). Then \(x^2-11x+18=(x-\tfrac{11}{2})^2-\tfrac{49}{4}\).

What is the vertex of \(y=x^2-11x+18\)?

Vertex \(x\)-coordinate is \(h=\tfrac{11}{2}\). Then \(y=\left(\tfrac{11}{2}\right)^2-11\cdot\tfrac{11}{2}+18=-\tfrac{49}{4}\).

What are the \(x\)-intercepts and \(y\)-intercept?

\(x\)-intercepts: \(x=2,9\). \(y\)-intercept: set \(x=0\), so \(y=18\).

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