Q. \(x^2 – 9x + 20\)

Q: Factor x^2-9x+20.

\left(x-4\right)\left(x-5\right).

Q: Find the roots of x^2-9x+20 using the quadratic formula.

x=\dfrac{9\pm\sqrt{81-80}}{2}=\dfrac{9\pm1}{2}, so x=4,5.

Q: What are the y-intercept and does the parabola cross it?

Intercept at x=0: 20. Since 20>0 and roots are 4,5, it does cross the x-axis at those roots, not at the intercept.

Q: Find the vertex of y=x^2-9x+20.

x-coordinate =\dfrac{9}{2}=4.5. y= (4.5)^2-9(4.5)+20=0.25, so vertex is (4.5,0.25).

Q: Write the function in completed-square form.

x^2-9x+20=\left(x-\dfrac{9}{2}\right)^2+\dfrac{1}{4}.

Q: Determine the axis of symmetry and whether it opens up or down.

Axis x=\dfrac{9}{2}. Since leading coefficient is positive, it opens up.

Answer

We factor the quadratic by finding two numbers that multiply to \(20\) and add to \(-9\), which are \(-5\) and \(-4\).

\[
x^2-9x+20=(x-5)(x-4)
\]

So the factored form is \((x-5)(x-4)\).

Detailed Explanation

We want to factor the quadratic expression \(x^2 – 9x + 20\).

Step 1: Identify the quadratic coefficients.

The expression is \(x^2 – 9x + 20\).

So the coefficients are:

\(a = 1\)
\(b = -9\)
\(c = 20\)

Step 2: Find two numbers that multiply to \(ac\) and add to \(b\).

Compute \(ac\):

\[
ac = 1 \cdot 20 = 20
\]

Now we need two integers such that:

their product is \(20\)
their sum is \(-9\)

List factor pairs of \(20\):

\(1\) and \(20\), sum \(21\)
\(-1\) and \(-20\), sum \(-21\)
\(2\) and \(10\), sum \(12\)
\(-2\) and \(-10\), sum \(-12\)
\(4\) and \(5\), sum \(9\)
\(-4\) and \(-5\), sum \(-9\)

The pair \(-4\) and \(-5\) works because:

\((-4)(-5) = 20\)
\(-4 + (-5) = -9\)

Step 3: Write the factored form.

Using the numbers \(-4\) and \(-5\), the quadratic factors as:

\[
x^2 – 9x + 20 = (x – 4)(x – 5)
\]

Final Answer:

\[
\boxed{x^2 – 9x + 20 = (x – 4)(x – 5)}
\]

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Algebra FAQ

Factor \(x^2-9x+20\).

\(\left(x-4\right)\left(x-5\right)\).

Solve \(x^2-9x+20=0\).

From factoring, \(x=4\) or \(x=5\).

Find the roots of \(x^2-9x+20\) using the quadratic formula.

\(x=\dfrac{9\pm\sqrt{81-80}}{2}=\dfrac{9\pm1}{2}\), so \(x=4,5\).

What are the \(y\)-intercept and does the parabola cross it?

Intercept at \(x=0\): \(20\). Since \(20>0\) and roots are \(4,5\), it does cross the \(x\)-axis at those roots, not at the intercept.

Find the vertex of \(y=x^2-9x+20\).

\(x\)-coordinate \(=\dfrac{9}{2}=4.5\). \(y= (4.5)^2-9(4.5)+20=0.25\), so vertex is \((4.5,0.25)\).

Write the function in completed-square form.

\(x^2-9x+20=\left(x-\dfrac{9}{2}\right)^2+\dfrac{1}{4}\).

Determine the axis of symmetry and whether it opens up or down.

Axis \(x=\dfrac{9}{2}\). Since leading coefficient is positive, it opens up.

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