Question

cc.2.2.hs.d.10
solve $x^2 - 3x - 18 = 0$ by factoring.
graph to verify your solutions.
\bigcirc \\{6\\}
\bigcirc \\{-6, 3\\}
\bigcirc \\{-9, 2\\}
\bigcirc \\{-3, 6\\}
question 5
4 pts
cc.2.2.hs.d.4
which quadratic equation has roots -2 and 4?

Explanation:

First Question (Solving \(x^2 - 3x - 18 = 0\) by factoring)

Step1: Factor the quadratic

We need two numbers that multiply to \(-18\) and add to \(-3\). The numbers are \(-6\) and \(3\) since \((-6)\times3=-18\) and \(-6 + 3=-3\). So, \(x^2 - 3x - 18=(x - 6)(x + 3)\).

Step2: Solve for \(x\)

Set each factor equal to zero:

• \(x - 6 = 0\) gives \(x = 6\)
• \(x + 3 = 0\) gives \(x=-3\)

So the solutions are \(x = -3\) and \(x = 6\), which is the set \(\{-3, 6\}\).

Second Question (Quadratic equation with roots \(-2\) and \(4\))

If a quadratic equation has roots \(r_1\) and \(r_2\), the equation can be written in factored form as \((x - r_1)(x - r_2)=0\). Here, \(r_1=-2\) and \(r_2 = 4\).

Step1: Substitute roots into factored form

\((x - (-2))(x - 4)=0\) which simplifies to \((x + 2)(x - 4)=0\)

Step2: Expand the factored form

Using the distributive property (FOIL method):
\(x\times x+x\times(-4)+2\times x + 2\times(-4)=0\)
\(x^2-4x + 2x-8 = 0\)
\(x^2-2x - 8 = 0\)

Answer:

s:

• For the first question: \(\{-3, 6\}\) (corresponding to the option \(\{-3, 6\}\))
• For the second question: The quadratic equation is \(x^2 - 2x - 8 = 0\)