SOLUTION: Find two consecutive integers so that twice the square of the larger is 57 more than three times the smaller. Find the integers.
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Question 169328This question is from textbook Integrated Arithmetic and Basic Algebra
: Find two consecutive integers so that twice the square of the larger is 57 more than three times the smaller. Find the integers.
This question is from textbook Integrated Arithmetic and Basic Algebra
Answer by checkley77(12844) (Show Source): You can put this solution on YOUR website!
Let x & x+1 be the 2 integers.
2(x+1)^2=3x+57
2(x^2+2x+1)=3x+57
2x^2+4x+2=3x+57
2x^2+4x-3x-57+2=0
2x^2+x-55=0
(2x+11)(x-5)=0
2x=11=0
2x=-11
x=-11/2
x=-5.5 ans. -5.5+1=-4.5 ans.
x-5=0
x=5 ans. 5+1=6 ans.
Proof:
2*6^2=3*5+57
2*36=15+57
72=72
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