Question

5x^2 + 11x + 2

Answer 1

You have a quadratic expression: \(5x^2 + 11x + 2\). Here are a few things we can do with this expression:

1. **Factor the expression (if possible)**:
We look for two numbers that multiply to \(5 \times 2 = 10\) (the product of the coefficient of \(x^2\) and the constant term) and add to \(11\) (the coefficient of \(x\)). The pair \(10\) and \(1\) satisfies this condition.

So we can rewrite the middle term:
\[
5x^2 + 10x + 1x + 2
\]
Now, we can factor by grouping:
\[
(5x^2 + 10x) + (1x + 2) = 5x(x + 2) + 1(x + 2)
\]
This gives us:
\[
(5x + 1)(x + 2)
\]

2. **Finding the roots (zeros) of the quadratic**:
We can use the quadratic formula, which is given by:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \(a = 5\), \(b = 11\), and \(c = 2\).

First, we calculate the discriminant:
\[
b^2 - 4ac = 11^2 - 4(5)(2) = 121 - 40 = 81
\]
Now, substituting into the quadratic formula:
\[
x = \frac{-11 \pm \sqrt{81}}{2(5)} = \frac{-11 \pm 9}{10}
\]
This gives us two roots:
\[
x_1 = \frac{-2}{10} = -0.2 \quad \text{and} \quad x_2 = \frac{-20}{10} = -2
\]

3. **Graphing the quadratic**:
If you were to graph this quadratic function, it would be a parabola opening upwards (since the coefficient of \(x^2\) is positive). The vertex can be found using the vertex formula:
\[
x = -\frac{b}{2a} = -\frac{11}{2 \cdot 5} = -\frac{11}{10} = -1.1
\]
You can substitute this back into the expression to find the y-coordinate of the vertex.

If you need more specific information or help with a particular aspect, feel free to ask!