 In this video, we provide the solution to question number six from practice exam number two from math 1050 in which case we're given three quadratic equations and asked which of these equations have at least one real solution so they could have one solution or two solutions in particular we want to rule out any that have to non real complex solutions and we don't have to solve the equations we just have to recognize the nature of the solution set so I'm going to try to use the discriminant to help me out here so remember the discriminant is this B squared minus 4ac it's the part that shows up inside the the square root of the quadratic form of the radicand there and so for the first one I write here your discriminant would look like the B of course is the linear coefficient there so 4 squared is 16 minus for the next one we're going to get 4 times 1 times negative 1 notice this is going to equal 20 when you simplify it that is greater than 0 so in this situation you would have two real solutions so we're going to check off I that's going to be part of our list what about to hear double I same basic idea we're going to compute the discriminant we're going to take a negative 12 squared that's going to give us 144 then we're going to subtract from it 4 times 4 times 9 like so for which you know continue with the calculation right here but you're going to actually quickly notice that 4 times 4 times 9 is actually 144 again in which case then you end up with 0 0 still okay that actually means you have exactly one real solution this in retrospect we could see is actually a perfect square trinomial this thing would factor as 2x minus 3 squared you don't need to know that but be aware that is a perfect square trinomial has exactly one real solution it's just the vertex of the parabola thus it's going to be including on our list but you notice here that you know we've already ruled out some of the possibilities since one and two are there you know d could be an answer but it can also be like g what about 3 let's check that one so for the last one here we'll compute the determinant so we're going to get negative 6 squared which is 36 minus 4 times 1 times 12 like so in that situation what do we end up with well 4 times 12 is 48 so you're going to get 36 minus 48 this already we already know is negative at this moment but if you want to finish it up you're going to get another 12 it's less than zero since the discriminant is negative that means there's no real solution so we don't get to therefore since one and two have two and one solutions respectively and three does not we see the correct answer be choice d one and two have real solutions number three has no real solutions