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# Quadratic equations word problem: triangle dimensions

## Video transcript

the height of a triangle is four inches less than the length of the base the area of the triangle is 30 inches squared find the height and base use the formula area equals one-half base times height for the area of a triangle okay so let's think about it a little bit we have the bit let me draw a triangle here so this is our triangle and let's say that the length of this bottom side that's the base let's call that B and then this is the height this is the height right over here and then the area is equal to 1/2 base times side area is equal to 1/2 base times height now in this first sentence they tell us that the height of the height of a triangle is 4 inches less than the length of the base so the height is equal to the base minus 4 that's what that first sentence tells us the area of the triangle is 30 inches squared so if we take 1/2 the base times the height we'll get 30 inches squared or we could say we could say that 30 inches squared is equal to 1/2 times the base times the base times the height now instead of putting an H in for height we know that the height is the same thing as 4 less than the base so let's put that in there for less than the base and then let's see what we get here we get let me do this in yellow we get 30 is equal to 1/2 times let's distribute this B times B let me make it clear actually let's let's do it this way times B over 2 times B minus 4 I just multiplied the 1/2 times the B now let's distribute the B over 2 so 30 is equal to B squared over over 2 be careful B over 2 times B is just B squared over 2 and then B over 2 times negative 4 is negative 2b now just to get rid of this fraction here let's multiply both sides of this equation by 2 so let's multiply that side by 2 and let's multiply that side by two on the left-hand side you get 60 on the right-hand side two times B squared over two is just B squared negative 2 B times 2 is negative 4 B and now we have a quadratic here and the best way to solve a quadratic we have a second degree term right here is to get all of our all of the terms on one side of the equation having them equal 0 so let's subtract 60 from both sides of this equation let's subtract 60 from both sides and we get 0 is equal to B squared minus 4 B minus 60 and so what we need to do here is just factor this thing right now or factor it and then know that if I have the product of some things and that equals 0 that means that either one or both of those things need to be equal to 0 so we need to factor B squared minus 4 B minus 60 so what we want to do we want to find two numbers whose sum is negative 4 whose sum is negative 4 and whose product is negative 60 so we want to find two numbers whose sum is equal to negative 4 and whose product is equal to negative 60 now given that their product is negative we know there are different signs and this tells us that they're their absolute values are going to be 4 apart that one is going to be 4 less than the other so you can look at the products of the factors of 60 1 and 60 are too far apart even if you made one of them negative you would either get positive 59 as a sum or negative 59 as a sum 2 and 30 is still too far apart 3 and 15 3 and 20 sorry it's still too far apart if you have made one negative you would either get negative 17 or positive 17 then you can have 4 and 15 still too far apart if you made one of them negative their sum would be either negative 11 or positive 11 then you have 5 and 12 still seems too far apart from each other one of them is negative then you would either have there's some being positive 7 or negative 7 then you have 6 and ten now this looks interesting therefore apart so if we make and we want the larger absolute magnitude number to be negative so that their sum is negative so if we make it six and negative 10 their sum will be negative 4 and their product is negative 60 so that works so you can literally say that this is equal to this is equal to B plus 6 times B minus 10 B plus 2 a times B minus or plus B minus the B and now let me be very careful here this B over here I want to make it very clear is different is different than the B that we're using in the equation I just use this B here to say look we're looking for two numbers that add up to this second term right over here it's a different B I could have said X plus y is equal to negative 4 and x times y is equal to negative 60 in fact let me do it that way just so we don't get confused so we could write x plus y is equal to negative 4 and then we have x times y is equal to negative 60 so we have B plus 6 times B plus y X is 6 Y is negative 10 and that is equal to 0 and you could you could well let's just solve this right here and then we'll go back and show you that you could also factor this by grouping but just from this we know that either one of these is equal to 0 either B plus 6 is equal to 0 or B minus 10 or B minus 10 is equal to 0 if we subtract 6 from both sides of this equation we get B is equal to negative 6 or if you add 10 to both sides of this equation you get B is equal to 10 and those are our two solutions you could put them back in and verify that they satisfy our constraints now the other way that you could solve this and we're going to get the exact same answer is you can just break up this negative 4b into its constituents so you could have broken this up into 0 is equal to B squared and then you could broken it up into +6 B minus 10 B plus 6 B minus 10 B minus 60 and then factor it by grouping group these first two terms group these second two terms just going to add them together the first one you can factor out a B so you have B times B plus 6 the second one you can factor out a negative 10 so minus 10 times B plus 6 all that's equal to 0 and now you can factor out a B plus 6 so if you factor out a B plus 6 here then you get 0 is equal to B minus 10 times B plus 6 we're literally just factoring out we're literally just factoring out this out of the expression you're just left with the B minus 10 B minus 10 you get the same thing that we did in one step over here whatever works for you but either way the solutions are either B is equal to negative 6 or B is equal to 10 and we have to be careful here remember this is a word problem we can't just stay Toby could be negative 6 or B could be 10 we have to we have to think about whether this makes sense in the context of the actual problem we're talking about lengths of triangles or lengths of the sides of triangles we can't have a negative length we can't have a negative length so because of that we can't the base of a triangle can't have lengths negative 6 so we can cross that out so we actually only have one solution here almost made a careless mistake forgot that we were dealing with a word problem the only possible base is 10 and let's see they say find the height and the base once again not done so the base we're saying is 10 the height is 4 inches less it's B minus 4 so the height is 6 and then you can verify the area is 6 times 10 times 1/2 which is 30