in this lesson we're going to talk about how to solve quadratic equations that contain imaginary Solutions so what you need to remember is that the Square < t of -1 is equal to I so let's keep that in mind let's start with the first equation on the left in order to solve it we need to get X by itself so let's begin by subtracting both sides by 16 so we're going to get x^2 is equal to -16 at this point we could take the square root of both sides theare < TK of x^2 is just X but we're going to get plus or minus < TK -16 now theare < TK of -16 we could write it as theare < TK of 16 * theare < TK of-1 the square root of 16 is 4 and the square OT of1 is I so the answer is plus or minus 4 I so what that means is that we have two imaginary Solutions the first solution is positive 4 I and the second solution is NE 4 I now let's move on to the next one in this example we're going to start off with the same procedure we're going to subtract both sides by 49 so we're going to have 4x^2 is = to -49 now before we take the square root of both sides let's divide both sides by four so right now we have x^2 is equal to - 494 now let's take the square root of both sides so we're going to have X on the left and then plus or minus now we can re write this square root as theare < TK of 49 / the < TK of 4 * the < TK of1 I just broke it up into three parts now the square root of 49 is 7 the square < TK of 4 is 2 and the square < TK of -1 is I so our two answers are positive 7 over2 I and 7 over2 I but you could just leave your answer like this x is equal to plus or minus 7 over 2 I now for this example instead of a binomial we have a quadratic equation that's a trinomial go ahead and find a solution to this one now we can't Factor this trinomial if you were to try to find two numbers that multiply to seven but add to three it's not going to work I mean you have 1 and seven at least you're not going to get any integers that will give you a solution here so we can't Factor this trinomial what we can do is we could use the quadratic equation which is X is equal to B plus or minus the square < TK of b^2 - 4 a c / by 2 a a now this quadratic equation is in the form ax + BX + C is equal to Z so that means a is 1 B is 3 C is 7 so using this formula it's going to be -3 plus orus < TK b^2 which is 3^ 2 - 4 A C or 4 * 1 * 7 / 2 a or 2 * 1 so right now we need to simplify this expression 3^ 2 is 3 * 3 which is 9 4 * 7 is 28 9 - 28 is -19 now at this point we could break up the fraction into two smaller fractions so the first part is going to be 3 over2 or rather -3 over2 and then plus or minus the second part will be square < TK 9 /2 now the square < TK of -9 that's the < TK of 19 * theare < TK of1 which isun 19 * I so we can write that as plus or minus the < TK of 19 / 2 times the imagin number I so we could leave our answer in this format so notice that this is in a plus or minus bi i 4 format so that's the answer for this problem now let's try a similar example if you want to pause the video feel free to work on this problem so once again we have a quadratic equation in the form of a trinomial so a is 3 B is 5 C is 10 so let's use the quadratic formula to get the solution by the way if you want to see if you can Factor this trinomial what we need to do is multiply the leading coefficient by the constant term which will be 3 * 10 which is 30 and then we need to find two numbers that multiply to 30 but add to five so this would be 1 and 30 2 and 15 3 and 10 five and six none of these pairs will add or subtract to five so we can't Factor it we have to use the quadratic formula but that's how you could determine if you can Factor it for future problems so in this example B is 5 so we have -5 plus or minus the square < TK of 5^ s - 4 a c a is 3 C is 10 / 2 a or 2 * 3 5 * 5 is 25 4 * 3 is 12 * 10 that's 120 and 2 * 3 is 6 now 25 minus 120 that's going to be 95 the square root of 95 95 is 5 * 19 and we can't take the square root of either so it's best to leave it as 95 * the < TK of1 which is the < TK of 95 * I so the answer is going to be -5 over 6 plus or minus the squ < TK of 95 over 6 * I we just got to break it up into two fractions so that we can write our final answer in a plus or minus bi format so that's a standard format when dealing with complex numbers or imaginary numbers so that's basically it that's how you can solve a quadratic equation that contains imaginary Solutions so all you got to do is remember that the square root of negative 1 is I so if you have a negative sign within an even within a square root with an even index number like 2 4 six you're dealing with imaginary numbers but most of the times you won't see like four or six this is typically a two