Biancakuyukiviz
Answered

find the three consecutive multiple of five so that the square of the third, decreased by five times the second number, is the same as twenty five more than the twice the product of the first two numbers

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Mlcparra16viz
Let the second number be x

[tex] (x+5)^{2}-5x=2[(x-5)x]+25 [/tex]
[tex] x^{2} +10x+25-5x=2[ x^{2} -5x]+25[/tex]
[tex] x^{2} +5x+25=2 x^{2} -10x+25[/tex]
[tex] x^{2} +5x=2 x^{2} -10x[/tex]
[tex]5x= x^{2} -10x[/tex]
[tex]15x= x^{2} [/tex]
[tex]15=x[/tex]

Therefore the three multiples of 5 are 10, 15 and 20