SOLUTION: for each equation, determine what type of number the solutions are and how many solutions exists. x^2+10x+11=0 i've tried to do it with the quadratic equation but im not sur

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: for each equation, determine what type of number the solutions are and how many solutions exists. x^2+10x+11=0 i've tried to do it with the quadratic equation but im not sur      Log On


   

Question 679859: for each equation, determine what type of number the solutions are and how many solutions exists.
x^2+10x+11=0
i've tried to do it with the quadratic equation but im not sure if i have to use the quadratic equation.
Found 2 solutions by ReadingBoosters, josmiceli:
Answer by ReadingBoosters(3246) About Me  (Show Source):
You can put this solution on YOUR website!
A polynomial to the nth degree has n roots.
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This equation will hae 2 roots/solutions.
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The solutions are irrational.
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Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B10x%2B11+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%2810%29%5E2-4%2A1%2A11=56.

Discriminant d=56 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-10%2B-sqrt%28+56+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%2810%29%2Bsqrt%28+56+%29%29%2F2%5C1+=+-1.25834261322606
x%5B2%5D+=+%28-%2810%29-sqrt%28+56+%29%29%2F2%5C1+=+-8.74165738677394

Quadratic expression 1x%5E2%2B10x%2B11 can be factored:
1x%5E2%2B10x%2B11+=+1%28x--1.25834261322606%29%2A%28x--8.74165738677394%29
Again, the answer is: -1.25834261322606, -8.74165738677394. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B10%2Ax%2B11+%29

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Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
For what they are asking, you just need the part of the
quadratic formula called the " discriminant ". It is
+b%5E2+-+4%2Aa%2Ac+ which is inside a square root sign.
The rules are:
If +b%5E2+-+4%2Aa%2Ac+=+0+, 1 real solution
If +b%5E2+-+4%2Aa%2Ac+%3C+0+, 2 imaginary solutions
If +b%5E2+-+4%2Aa%2Ac+%3E+0+, 2 real solutions
------------------------------------
That's all you need for this problem.
+x%5E2+%2B10x+%2B11=0+
+a+=+1+
+b+=+10+
+c+=+11+
+b%5E2+-+4%2Aa%2Ac+=+10%5E2+-+4%2A1%2A11+
+b%5E2+-+4%2Aa%2Ac+=+100+-+44+
+b%5E2+-+4%2Aa%2Ac+=+56+
This is greater than zero, so there are 2 real solutions
Here's a plot of the equation:
+graph%28+400%2C+400%2C+-10%2C+10%2C+-16%2C+10%2C+x%5E2+%2B+10x+%2B+11+%29+
You can see that the solutions are (2) real negative numbers
You can get the fact they are negative from the rest of the quadratic formula:
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+