SOLUTION: Find two consecutive integer such that the squares of the larger diminished by five times the smaller is equal to 71.
Algebra.Com
Question 1077709: Find two consecutive integer such that the squares of the larger diminished by five times the smaller is equal to 71.
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
x and x+1
(x+1)^2-5x=71
x^2+2x+1-5x=71
x^2-3x+1=71
x^2-3x-70=0
(x-10)(x+7)=0
x=10,-7
10 and 11
121-50=71, they work
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-7 and -6
36-(-35)=71, they work
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