Question

What are the x intercepts of f(x)=x^2+7x+2 ?

Answer 1

For this case we have a function of the form:

$y = f (x)$

Where:

$f (x) = x ^ 2 + 7x + 2$

To find the intersections with the x-axis, we make $y = 0$, that is:

$x ^ 2 + 7x + 2 = 0$

Where:

$a = 1\\b = 7\\c = 2$

To find the solution we apply the following formula:

$x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}$

Substituting:

$x = \frac {-7 \pm \sqrt {7 ^ 2-4 (1) (2)}} {2 (1)}\\x = \frac {-7 \pm \sqrt {49-8}} {2}\\x = \frac {-7 \pm \sqrt {41}} {2}$

We have two roots:

$x_ {1} = \frac {-7+ \sqrt {41}} {2}\\x_ {2} = \frac {-7- \sqrt {41}} {2}$

Thus, the intersections with the "x" axis are:

$(\frac {-7+ \sqrt {41}} {2}, 0)\\(\frac {-7- \sqrt {41}} {2}, 0)$

Answer:

$(\frac {-7+ \sqrt {41}} {2}, 0)\\(\frac {-7- \sqrt {41}} {2}, 0)$