SOLUTION: Find the x-intercepts y = x^2 + 5x + 2 How do I find the x intercept?

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Question 69482: Find the x-intercepts y = x^2 + 5x + 2
How do I find the x intercept?
Found 2 solutions by ankor@dixie-net.com, bucky:
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Find the x-intercepts y = x^2 + 5x + 2
:
The x intercepts occur when y = 0 so just write it as follows and find x:
:
x^2 + 5x + 2 = 0
:
This cannot be easily factored (no integer solutions) so use the quadratic formula
:
in the form ax^2 + bx + c: a=1; b=5 and c=2
:
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
:
Substitute for a, b, & c
x+=+%28-5+%2B-+sqrt%28+5%5E2-4%2A1%2A2+%29%29%2F%282%2A1%29+
:
x+=+%28-5+%2B-+sqrt%28+25+-+8+%29%29%2F%282%29+
:
The solutions are:
x+=+%28-5+%2B+sqrt%28+17+%29%29%2F%282%29+
and
x+=+%28-5+-+sqrt%28+17+%29%29%2F%282%29+
:
In decimals it would be x = -4.561553 and x = -0.4384472
:
A graphical presentation of the x intercepts:
+graph%28+300%2C+200%2C+-6%2C+4%2C+-8%2C+8%2C+x%5E2+%2B+5x+%2B+2%29+
:
Do you understand this now?

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Find the x-intercepts y = x^2 + 5x + 2
One way to do this is to graph the equation by selecting values of x, plugging them in, and finding corresponding values of y. Then plot the (x,y) points. This might help you to narrow down the places where the graph crosses the x-axis, but it also involve quite a lot of work if the answer is not a "nice" integer. [In this problem the answers are not integers.]
Think in terms of the axes of a graph. In order for a number to be on the x-axis, what does its y value have to be? Can you picture that y must be zero. For example, is the point (3, 4) on the x-axis? No, it is not because the y value of 4 tells you the point is located 4 units above the x-axis. A point such as (5, 0) is located on the x-axis because it is 0 units above the x-axis.
So to find if the equation has values on the x-axis you can set the value of y equal to zero and see if there are any values of x that will make that equation true. Setting y equal to zero results in the equation:
x%5E2+%2B+5x+%2B+2+=+0
[Notice that I have switched the equation around -- putting the y value (zero) on the right side and the polynomial in x on the left side. No reason for it, its just a little more conventional.]
If you could factor the equation to get it in the form
%28x+%2B-+a%29%2A%28x+%2B-+b%29+=+0
then you could set each of the factors equal to 0 because if either factor is zero the equation will be true. This would tell you the values of x for which the equation is true. However, in this problem the polynomial in x is not easily factored.
A more general method to solving this equation is to use the quadratic formula which says that in a quadratic equation of the standard form:
ax%5E2+%2B+bx+%2B+c+=+0
the values of x that satisfy this equation are:
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
By comparing the equation from this problem to the standard form you can see that a = 1, b = +5, and c = +2. Plugging these values into the quadratic formula tells you that the two values of x that satisfy your problem are:
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x+=+%28-5+%2B-+sqrt%28+5%5E2-4%2A1%2A2+%29%29%2F%282%2A1%29+
and this can be simplified to:
x+=+%28-5+%2B-+sqrt%28+25-8+%29%29%2F%282%29+
which can be further simplified to:
x+=+%28-5+%2B-+sqrt%2817%29%29%2F%282%29
as an approximation you can say that the square root of 17 is about 4.1231 so that the two values for x are:
x+=+%28-5+%2B+4.1231%29%2F2
and
x+=+%28-5+-+4.1231%29%2F2
These simplify to x=-0.43845 and x=-4.56155
However, these are only approximately correct because the square root of 17 is not a "nice" number.
Hope this helps.