Laura I. answered • 11/04/15
Tutor
New to Wyzant
Patient Tutor for Math and Science
Alejandro, I think the best thing to do is to play with the radicand, b2 -4ac because that is what ultimately makes the solutions be imaginary or not.
Let's start with having no real solutions. In order for that to happen, you would need a negative number inside the radical. This means that when subtracted, b2 must be smaller than 4ac (b2<4ac). Pick any number for a, b, and c but make sure that when b is squared it is larger than when 4 is multiplied with a and c.
Example: a=2, b=6, c=10
x= -b±√(b2-4ac) /2a
x= -6±√(62-4•2•10) / (2•1)
x= -6±√(36-80) / 2
x= -6±√(-44) /2
You can't take a negative square root so there are no real solutions.
For having only one real solution, think about how that can happen. When the problem is at the final stages of finding solutions, you would have to make sure that there would be no difference. Ultimately, this means that the radicand is somehow cancelled out. In other words, b2 would have to equal 4ac (b2=4ac). Again, pick any number for a, b, and c but make sure that when subtracted the randicand is 0.
Example: a=1, b=2, c=1
x= -b±√(b2-4ac) /2a
x= -2±√(22-4•1•1) / (2•1)
x= -2±√(4-4) /2
x=-2±√0 /2
x= (-2 + √0)/2 x= (-2 - √0)/2
√0 is 0 so it doesn't matter if it's added or subtracted with another number. It will cancel.
x= -2/2 or -1
As for having two distinctive solutions, it's easier to have the radicand be a square number. For this, try to pick square number for b and have 4ac equal a smaller number so when subtracted, you get a square. Near the end you will see that you will have two noticeably different solutions!
Example: a=3, b=10, c=3
x= -2±√(b2-4ac) /2a
x= -10±√(102 -4•3•3) / (2•3)
x= -10±√(100-36) /6
x= -10±√(64) /6
x= -10±8 /6
x= (-10 +8)/6 x= (-10 - 8)/6
x= -2/6 x= -18/6
x= -1/3 x= -3
There are two different answers!
I hope this helps!
