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

User Priyatham
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1 Answer

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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)

User Daphshez
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