In Exercises 115–122, find all values of x satisfying the given c...

Question

In Exercises 115–122, find all values of x satisfying the given conditions.

y1 = 2x^2 + 5x - 4, y2 = - x^2 + 15x - 10, and y1 - y2 = 0

Accepted Answer

Hello today we're going to be solving for X. Using the given conditions. So the first condition given to us is Z one is equal to X. And the second condition is Z two is equal to negative six over X. And we are told that Z one plus Z two is equal to negative one. So what we're going to go ahead and do is use this statement to solve for X. And in order to use this we're going to take our given Z one and Z two and plug it in for this Z one and Z two. So doing so is going to give me Z one which is X plus Z two which is negative six over X. And that's going to be equal to negative one. Now opening up this parentheses E this is going to give me x minus six over X is equal to negative one. And now we need to go ahead and sell for X. Well since I want to get X out of this denominator, I'm going to go ahead and isolate this quantity first and in order to do that, I'm going to subtract X to both sides of the equation. Doing so is going to give me negative six over X equal to negative X minus one. And if I multiply both sides of the equation by negative X. That's going to give me positive six is equal to negative x times the quantity negative X -1. Now I need to go ahead and distribute negative X into this quantity and doing so is going to give me six is equal to negative X times negative X which is positive X squared times negative X times negative one which is going to be positive X. Now since we want to solve for X, we're gonna go ahead and set this equation equal to zero and we're going to do that by subtracting six to both sides of the equation. Doing so is going to give me X squared plus X minus six is equal to zero and this is a fact herbal trin Amiel as this try no meal is going to factor into the quantities X minus two times X plus three is equal to zero. And we're going to use our zero property to set both of these quantities equal to zero. Doing so is going to give me x minus two is equal to zero and X plus three is equal to zero. And we need to solve for X in both of these cases while on the left hand side I just need to add two to both sides of the equation. And doing so is going to give me X is equal to two which is our first solution to the given conditions. And on the other side I just need to go ahead and subtract three to both sides of the equation. And doing so is going to give me X is equal to negative three which is our second solution. Now we want to go ahead and represent these two solutions as a solution set well, in order to do that, we go ahead and write down our braces here and put the smallest number on the left hand side, which is negative three. Then the biggest number goes on the right hand side, which is two, and you close it with your other brace. And this is going to be the solution set to these given conditions. 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 problem. And I'll go ahead and see you all in the next video.

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In Exercises 115–122, find all values of x satisfying the given conditions. y1 = 2x^2 + 5x - 4, y2 = - x^2 + 15x - 10, and y1 - y2 = 0