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 = 4x - 7

Accepted Answer

Hello today we're going to be solving the given equation using any of our preferred methods. So the equation given to us is x squared is equal to six X minus 11. Now in order to solve this equation I need to go ahead and set the equation equal to zero. And in order to do that I can subtract six X from both sides of the equation and add 11 to both sides. Doing so is going to give me x squared minus six. X plus 11 is equal to zero and we can go ahead and solve this using the quadratic equation. So let's just quickly recall what the quadratic equation states. It states that X is equal to negative B plus or minus the square root of B squared minus four times a times C. All of that over two times A and a B C are going to be the leading coefficients in your given try no meal. So for this case here A is going to be the coefficient in front of our x square term which here it's going to be one. B. Is the coefficient in front of our X term which is negative six. And C is just gonna be the constant at the end which is positive 11. Now we just need to go ahead and plug this into the quadratic equation. Doing so is going to give me X is equal to negative B which is negative negative six. So positive six plus or minus the square root of negative six squared minus four times A which is one times C which is 11. All of that over two times a which is two times one. Now let's go ahead and simplify negative six squared is going to give us 36 negative four times one times 11 is going to give us 44. So we have X is equal to six plus or minus the square root of 36 minus 44. All over two times one which is two. Now 36 minus 44 is going to give us negative eight, so X is going to equal to six plus or minus the square root of negative 8/2. Now recall that you are not allowed to have a negative number inside of a square root. So whenever you have a negative number inside the square root it produces an imaginary number. So for instance here the square root of negative X can be represented as the square root of X times I where I is the imaginary number. So in this case here we can go ahead and rewrite negative eight as the square root of a I. So X is going to equal to six plus or minus the square root of eight. I over to. Now the square root of eight can be represented as two times the square root of two, so X is equal to six plus or minus two times the square root of two. I over two. And if we take a look at our leading coefficients in the numerator, they both have a common factor of two. So we can factor out a two from the lead from the numerator. Doing so is going to give me X is equal to two times three plus or minus the square root of two, I over two. And since we have a two in the numerator and denominator, we can go ahead and cancel them out. And that leaves us with X is equal to three plus or minus the square root of two. I now we can't simplify this any further because this is a complex number, but we can go ahead and separate it because it represents a positive and negative value. So X is going to equal to three plus the square root of two, I and X is equal to three minus the square root of two. I. And these are going to be the solutions to this equation. And with that being said, the solution to this problem is going to be a 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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