Solve each equation in Exercises 65–74 using the quadratic formul...

Question

Solve each equation in Exercises 65–74 using the quadratic formula.

x^2 - 6x + 10 = 0

Accepted Answer

Hello today we're going to be solving the following equation using the quadratic formula. So the quadratic formula is when x is equal to negative B plus or minus the square root of B squared minus four times a times C. All over two times A. And the equation given to us is x squared minus 16, X plus 73 is equal to zero. So in order to use the quadratic equation, we need to go ahead and figure out what our abc are going to be. And those values are just going to be the leading coefficients in our given equation. So A is going to be the coefficient in front of X squared which is just one. So A is equal to one. B is the coefficient in front of our X term. So B is equal to negative 16 and C is just the last coefficient. So C is going to equal to 73. Now what we need to do is go ahead and plug this into our quadratic formula. So X is going to equal to negative B which is negative negative 16. So positive 16 plus or minus the square root of negative 16 squared minus four times A which is one times C which is 73 All over two times 1 Which is two times 8. Now, what we need to do is just go ahead and simplify negative 16 squared is going to give us 256 and negative four times one times 73 is going to give us negative 292. So what we have is X is equal to plus or minus the square root of -292. All over two times 1, which is two now, 256 minus 292 is going to give us the value of negative 36 so X is equal to 16 plus or minus the square root of negative 36/2. Now the problem here is that we have a negative number inside the square root while we have a way of representing negative numbers inside the square root, suppose that you have the square root of negative X. Since we're not allowed to have a negative number inside the square root, it produces an imaginary value. So the square root of negative X can be rewritten as the square root of X. I where I represents the imaginary number and we're gonna do the same thing here for negative 36. So the square root of negative 36 can be rewritten as the square root of 36. I So X is equal to 16 plus or minus the square root of 36. I all of that over to. Now we can go ahead and evaluate the square root of 36 because that is just going to give us six. So we have X is equal to 16 plus or minus six. I over two. Now in the numerator we have 16 and we have six so we can factor out a common factor of two. Doing so is going to give us X is equal to two times eight plus or minus three. I all over two. And since we have a two in the numerator and denominator, we can go ahead and cancel it out. Leaving us with X is equal to eight plus or minus three. I now we can go ahead and separate the values of X and get our two solutions. So X is equal to eight plus three. I and X is equal to eight minus three I. And this is going to be the solution to our equation. And with that being said, the solution to our problem is going to be a So I hope this video helps you in understanding how to solve using the quadratic formula and I'll go ahead and see you all in the next video.

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