To see how to solve an equation that involves the absolute value ...

Question

To see how to solve an equation that involves the absolute value of a quadratic polynomial, such as | x^2 - x | = 6, work Exercises 83–86 in order. For x^2 - x to have an absolute value equal to 6, what are the two possible values that x may assume? (Hint: One is positive and the other is negative.)

Accepted Answer

Hey everyone here we are asked to find the two possible values of X. For the expression X squared minus X to have an absolute value equal to 12. So from our given problem statement, we see we are trying to find the possible X values from the absolute value of the expression X squared minus X is equal to 12. So now with this equate we need to solve for X. So we first need to recall the property where if the absolute value of X is equal to A and that A is greater than zero, then we know that X must be equal to A. Or X is equal to negative A. So now with this property here we can go ahead and set our expression X squared minus X equal to both positive 12 and negative 12. So we can start with positive 12 And solving for X. We first just want to move this 12 to the left hand side by subtracting 12 from both sides of the equation. So we have x squared minus x minus 12 is equal to zero. Now to solve for X, we just need to factor this expression on the left hand side and we see we have the equal of x minus four times the quantity of X plus three is equal to zero. Now from this expression on the left hand side we need to find the X. Values that make this entire expression equal to zero. So that means we can set each individual part of the expression equal to zero. So we can start with the first set of parentheses and we have x minus four again setting it equal to z zero. Now we can add four to both sides to get X alone. And we see that X is equal to four. Now from the other part of our expression X plus three we can set it also equal to zero. Now subtracting three from both sides to get X alone, we see that X is also equal to negative three. So here we have found two values from setting our original expression equal to positive 12. Well now we need to set the same expression equal to negative 12. So we have X squared minus X. Is equal to negative 12. Now again we need to start solving for X. So we first want to move this negative 12 to the left hand side. So we need to add 12 to both sides of the equation and we have x squared minus X plus 12 is equal to zero. Now before we start trying to factor this expression to solve for X, we first want to determine if we can get a real solution from factoring this expression. So we first need to find the discriminate. Well if we recall the formula for finding the discriminate we have D. Is equal to b squared minus for a C. And now using this formula we can find the discriminate for our expression that we have here. So we have D. Is equal to B squared which is going to be one negative one squared the quantity of negative one squared minus four times A, which is one times C, which is 12. And now we see after simplifying that our discriminate is equal to negative 47. And since our discriminate is a negative, there is no real solution from setting our expression equal to negative 12. Therefore the original X. Values. We found our our final answer. So again originally we found that X is equal to four and X is equal to negative three. And this is our final answer here which is answer choice. A thanks for watching. I hope you found this video helpful and I'll see you again next time.

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