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In order to build a deck at a right angle to their house, Lucinda and Felipe place a stake in the ground a precise distance from the back wall of their house. This stake will combine with two marks on the house to form a right triangle. From a course in geometry, Lucinda remembers that there are three consecutive integers that can work as sides of a right triangle. Find the sides of that triangle.

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Missouri State University

McMaster University

Harvey Mudd College

Numerade Educator

So this is a word problem involving a right angle. The word problem itself. After you read through it outlines the looks of the right triangle that is to be created. So when reading the word problem, you are given a leg right here that will be called Axe. Consecutive integers will create this triangle and consecutive integers we know. 34567 those air consecutive. Each number is one grader than the number before it. So there's the first consecutive integer after acts, and then the next consecutive integer after acts will be an additional one after and there then becomes my high pot news. How do I know that's my hypothesis? The green one. I know that because it is the larger of the three numbers. Good practice wrap any by no meals in parentheses. Whenever you are going to have to square those, it becomes much more complex than it looks. My right angle is right here, so the question is, what are the dimensions of the three legs? So we will have three answers. When we're finished, let us go to Pythagorean steering, which states that the square some, which is in addition, sign of the square of the two legs which we will call and be will always be equal to the square of the hip pot news which is always see now a and B don't matter which order you put in the two legs. Access one leg X plus one is the second leg. See? Does matter. It must be the X plus two. So when I read left to right, the first leg I see is the axe. So I'm gonna substitute the accent for the A and I get X squared. I'm gonna substitute in parentheses, acts plus one and for B and I will get the by no meal square. This is not as simple as squaring the X and squaring the one and calling it X squared plus one. It is not that simple. So please remember that we'll get to the foiling of that. Just a moment. See, is X plus two again in parentheses. This is not as simple as X squared, plus four. Okay, Now my job is to solve for X but I first must remove all the parentheses and to remove the parentheses. I have to square the binomial. So the first term remains X square. I'm gonna write out in red everything that comes out of the X Plus one. We get the X plus I'm sorry, The x Times X, which is X square plus x times one which is one X times one plus acts, which is another one X time plus one times one that is X plus one squared, right, Therefore terms. Now I'm gonna do the same to my ax plus two x times X is X squared ext. Times two is two x que times x is two x and two times two this for there you go. So that is my by no meals squared Being very careful with the squaring off them. Now what I'm going to do is combine like terms on either side. What I do see is X squared plus X squared can be combined to become two X squared. Addition means you and the coefficients. You don't mess with the exponents. I see another set. A combine herbal like terms one x plus one x again and the coefficient one plus one is two. Don't mess with the exponents with e x. This one is flying solo right now. go to the right side. I see hopes. I'm sorry. I do not see multiple X squared, so that should be erased. That blue right there. Sorry about that. Got a little ahead of myself. What I do see are too linear Axis a two X and another two x which becomes a four x So the X squared I am so two x and two x four x and the floor four. I'm sorry is by himself as well. All right, now that all my life terms air put together, it's time to solve this through factoring. These are probably no meals. It's a quadratic X squared the square. The exponents determines that this is a quadratic and it will be a parable. A If it is graft, how do we solve quadratic? So we bring all terms to one side of the equation and the other side has to be a zero. And then we begin our factoring process. Now my role s Thomas. Which side you make zero keep your X squared positive, if at all possible. So if keeping in positive means take the left side to join the right side. Do that in this case taking the left side to join the right side is what's going to keep this positive. My ex word positive. It'll be a walk two X squared, minus one x squared. So that'll keep my ex word positive. So I'm gonna choose to make the right sides here. I'm gonna take away the X squared make that is your I'm gonna take away for acts. Make that a zero on the right, and then I'm going to take away the four and make that a zero. So every term on the right side just became a zero. Now, what do we do with that? We make we clean up the right side. Left side. I'm sorry. These all became zeros. Beautiful. That's what we need in order to factor. Now I'll combine two x squared minus X squared. And the fast thing you can be looking for is a positive one x squared. That will make your life the easiest it can possibly be When factory two x minus four X is minus two X one minus four is minus three. All right, so now that we have a positive one x squared, we can be assured that we will have to binomial factors equal to zero each beginning with an X because when I foil what I'm about to create, it will create an X squared. Now, what I have to do is think to myself what times What is negative? Three. What times? What is negative? Three. The ants too negative, too well, in my head, you start calculating and you realize negatives. Three and positive One is the answer to that question. So I'll put a minus three and a positive one negative. Three times positive. One is negative. Three negative three plus positive. One is negative. Two that is efficiently and effectively factored. Now we encounter the zero product property, the zero product property that says that when you have a series of factors equal to zero, all it takes is one of those factors to become a zero and you have solved the problem. So we have two factors, which means we're gonna have two possible solutions. Scenario one says factor one If factor one is equal to zero. You have just made the whole left side equal to zero. And you've made a true statement. So we will solve for this one right here. and get that acts is three. That is one of our solutions. Now there's also the other possible solution taking well. If the second factor equal zero, we have another solution that's valid and true. So we say factor to you can also equal zero. And then we saw for X and we see that access Negative one. Now we have a word problem. So what? We can't just circle these answers and call it good. The word problem is find the legs. Find the size of the triangle that will build this patio outside The house is negative. One a side of the triangle that's measurable. No, it is not. So. We will, in essence, say yes. That solution works, But not for the word problem. Will it work so well? Throw it out. We will use three. So right here we know that this side is going to be three cassettes. Where my access now what that x equal Street does for me to say anywhere I see Annex. Now I have the right to substitute a three and I will do that. Which tells me my rights. My red site is three plus one, which is four and my green side is three plus two, which is five, and we go back to the problem and look and see that we are talking. I do not see where it says feet or anything else like that. So we default to the word units. For these, it's five units, it's four units. It's three units. You should always check for labeling when you're doing a work problem.