SOLUTION: Please help me: 1. How many solutions exist for a quadratic equation? How do we detemine whether the solutions are real or complex?

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Question 154861: Please help me:
1. How many solutions exist for a quadratic equation? How do we detemine whether the solutions are real or complex?
Found 2 solutions by Fombitz, Earlsdon:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
How many solutions exist for a quadratic equation?
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The number of roots of a polynomial equation is equal to the degree of the polynomial (the exponent of the leading term).
Quadratic equations are of degree 2, x%5E2.
They have two (2) roots.
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How do we detemine whether the solutions are real or complex?
Use the discriminant.
For the general quadratic equation,
ax%5E2%2Bbx%2Bc=0
the discriminant is
D=b%5E2-4ac
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If D%3E0 then you have two distinct real roots.
Example:
x%5E2-7x%2B10=%28x-2%29%28x-5%29
D=49-4%2810%29=9
2 real roots, x=2,5 .
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If D=0, you have a double root, one real root occurring twice
x%5E2-2x%2B1=%28x-1%29%5E2
D=4-4%281%29=0
2 real roots, x=1,1.
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If D%3C0, you have two complex roots, that are complex conjugates.
x%5E2%2B1=%28x%2Bi%29%28x-i%29
D=0-4%281%29=-4
2 complex roots, x=i,-i.

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
In general, the number of solutions for a polynomial is equal to the degree of the polynomial.
A quadratic equation is a polynomial of degree 2 so it would have 2 solutions. The type of solutions a quadratic equation can be determined by examining the discriminant: %28b%5E2-4ac%29 which is taken from the quadratic formula:x+=+-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a
If the discriminant is negative, there are no real solutions/roots. This makes sense when you realize that a negative discriminant (the square root of a negative quantity) will yield complex solutions.
If the discriminant is zero, there is one real solution/root, sometimes referred to as a double root because you get two real solutions that are identical.
If the discriminant is positive, there are two real solutions/roots.
It is helpful to look at the graphs of quadratic equations with the above type of solutions/roots:
graph%28400%2C400%2C-5%2C5%2C-5%2C5%2C2x%5E2%2Bx%2B3%2Cx%5E2%2B6x%2B9%2Cx%5E2-5x%2B2%29
Green graph: y+=+2x%5E2%2Bx%2B3 Discriminant is negative, no real roots.
Red graph: y+=+x%5E2%2B6x%2B9 Discriminant is zero, one double root.
Blue graph: y+=+x%5E2-5x%2B2 Discriminant is positive, two real root.
As you can see, the roots or solutions to these equations are the x-values where the curves (parabolas) intersect the x-axis.