Q. \(x^2+9x+20\)

Q: Solve x^2+9x+20=0.

(x+4)(x+5)=0\Rightarrow x=-4,-5.

Q: What are the roots using the quadratic formula?

a=1,b=9,c=20. Discriminant b^2-4ac=81-80=1. So x=\frac{-9\pm1}{2}\Rightarrow x=-4,-5.

Q: Find the vertex (minimum) of y=x^2+9x+20.

Vertex at x=-\frac{b}{2a}=-\frac{9}{2}. Value y=\left(-\frac{9}{2}\right)^2+9\left(-\frac{9}{2}\right)+20=\frac14.

Q: What is the discriminant, and what does it mean here?

\Delta=b^2-4ac=1. Since \Delta>0, there are two distinct real roots, matching -4 and -5.

Q: Determine whether x^2+9x+20 is always positive.

Since the minimum value is \frac14>0, the expression is always positive for all real x.

Answer

Factor the quadratic by finding two numbers that add to \(9\) and multiply to \(20\): \(4\) and \(5\).

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

Detailed Explanation

We want to simplify and solve the expression

\[
x^2+9x+20.
\]

This is a quadratic expression. A common first step is to try factoring, because factoring makes the structure clear.

Step 1: Identify the coefficients

The expression is

\[
x^2 + 9x + 20,
\]
so the coefficients are:

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

Step 2: Factor the quadratic

We look for two numbers that:

Multiply to \(20\)
Add to \(9\)

Now list factor pairs of \(20\):

\(1 \cdot 20 = 20\) and \(1 + 20 = 21\)
\(2 \cdot 10 = 20\) and \(2 + 10 = 12\)
\(4 \cdot 5 = 20\) and \(4 + 5 = 9\)

Great: \(4\) and \(5\) multiply to \(20\) and add to \(9\). So we can factor:

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

Final Answer

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

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Algebra FAQ

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

\(x^2+9x+20=(x+4)(x+5)\).

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

\((x+4)(x+5)=0\Rightarrow x=-4,-5\).

What are the roots using the quadratic formula?

\(a=1,b=9,c=20\). Discriminant \(b^2-4ac=81-80=1\). So \(x=\frac{-9\pm1}{2}\Rightarrow x=-4,-5\).

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

Vertex at \(x=-\frac{b}{2a}=-\frac{9}{2}\). Value \(y=\left(-\frac{9}{2}\right)^2+9\left(-\frac{9}{2}\right)+20=\frac14\).

What is the discriminant, and what does it mean here?

\(\Delta=b^2-4ac=1\). Since \(\Delta>0\), there are two distinct real roots, matching \(-4\) and \(-5\).

Determine whether \(x^2+9x+20\) is always positive.

Since the minimum value is \(\frac14>0\), the expression is always positive for all real \(x\).

Solve x^2+9x+20 step by step.
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