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Question 953996: Find two consecutive positive integers such that the square of the second integer added to 7 times the first is equal to 253
Answer by algebrapro18(249)   (Show Source):
You can put this solution on YOUR website! For integers to be consecutive they come one after another(eg. 1,2 or 4,5 or 7,8). So if we let our first positive consecutive integer be x then our other consecutive integer is going to be x+1.
so we now that the square of the second integer added to 7 times the first is equal to 253. Writing this out using our consecutive integers we get:
(x+1)^2+7x=253
Now we can solve for our first consecutive integer.
(x+1)^2+7x=253 Expanding out the exponent
(x+1)(x+1)+7x=253 F.O.I.L.
x^2+x+x+1 +7x = 253 Combine like terms on the left hand side
x^2+9x+1 = 253 Subtract 253 from both sides to get into standard form
x^2+9x-252 = 0 Factor
I'm going to stop my work here and help you factor this. We need to think of numbers that will multiply to -252 but add to 9. So lets just list the factors of -252.
-1,252
-2,126
-3,84
-4,63
-6,42
-7,36
-9,28
-12,21
-14,18
Now which of these will add to be a positive 9?
-1+252 = 251
-2+126 = 124
-3+84 = 81
-4+63 = 59
-6+42 = 36
-7+36 = 29
-9+28 = 19
-12+21 = 9
-14+18 = 4
So we see that -12+21 =9 so our factorization is going to be:
(x-12)(x+21)=0
So now we can set each factor equal to zero and solve.
x-12 = 0 Add 12 to both sides
x = 12
x+21 = 0 Subtract 21 from both sides
x = -21
Now remember that our integers have to be positive since the problem asks about positive consecutive integers. So our first integer is going to be 12. So our next consecutive integer is 12+1 or 13.
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