SOLUTION: the sum of the squares of 3 consecutive numbers is 50. Let x be the smallest of the numbers. Show that 3x^2+6x+5=50
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Question 992500: the sum of the squares of 3 consecutive numbers is 50. Let x be the smallest of the numbers. Show that 3x^2+6x+5=50
Answer by Timnewman(323) (Show Source): You can put this solution on YOUR website!
Hi dear,
Let the numbers be x,x+1,x+2
Then, their squares are x²,(x+1)²,(x+2)²
If the sum of their squares is 90,
Then,
x²+(x+1)²+(x+2)²=90
Now expand the above,
x²+x²+2x+1+x²+4x+4=90
Then,
3x²+6x+5=90--->as required
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