Solve each equation in Exercises 83–108 by the method of your cho...

Question

Solve each equation in Exercises 83–108 by the method of your choice.

x^2 - 6x + 13 = 0

Accepted Answer

Hello today we're going to be solving the equation using any of our preferred methods. So the equation given to us is X squared plus two X plus five is equal to zero. And we're gonna go ahead and solve this using the quadratic formula. So the quadratic formula is 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. Now A B and C are just going to be the leading coefficients in our given Trina meal. So A. Is the coefficient in front of her X square term? Which in this case is one. So A. Is equal to one. B. Is gonna be the coefficient in front of her ex term. So B is equal to two and C is just gonna be the last coefficient. So C is equal to five. Now we're just gonna go ahead and take these three values and plug them into our quadratic equation. So X is equal to negative B which is negative two plus or minus the square root of two squared minus four times A which is one times C which is five all over two times A which is two times one. And now we just need to go ahead and simplify two squared is going to give us four and negative four times one times five is going to give us negative 20. So what we have is X is equal to negative two plus or minus the square root of four minus 20. All over two times 1 which is two Now 4 - is going to give us the value of -16. So X is equal to negative two plus or minus the square root of negative 16. All over two. Now the problem here is that we have a negative number inside the square root and you need to go ahead and represent that as an imaginary number since the negative numbers are not allowed inside square roots. So suppose you have the square root of negative X. This can be rewritten as the square root of X times I where I represents the imaginary number. So the square root of -16 can be rewritten as the square root of 16. I so X is equal to negative two plus or minus the square root of 16. I all over two and the square root of 16 I is equal to four, so X is equal to negative two plus or minus four. I all over two. Now in the numerator, both of our terms have a common factor of two. So we can go ahead and factor out two from the numerator and that gives us X is equal to two times negative one plus or minus two. I all over two. And since we have a two in both the numerator and denominator we can go ahead and cancel those out and that's going to leave us with X is equal to negative one plus or minus two. I now this is the most simplified form of our value of X. So the only thing we can do now is just go ahead and split up the values of X and that's going to be X is equal to negative one plus two, I and X is equal to negative one minus two I. And these are going to be the solutions to our given equation. And since these are the solutions to our equation, the answer to this problem is going to be D. So I hope this video helps you in understanding how to solve this type of equation. And I'll go ahead and see you all in the next video.

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Solve each equation in Exercises 83–108 by the method of your choice. x^2 - 6x + 13 = 0