out of three consecutive positive integers, the middle number is p. If three times the square of the largest is greater than sum of the other two numbers by 67. Calculate the value of p.
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I think the question is wrong in place of 67 it would be 68 than it will come a simple quadratic equation and then we can find it as
first let three consecutive no p-1,p,p+1 then
3(p+1)^2 = p-1+p +68
3p^2 +4p -64=0
thus p=4 by solving this
first let three consecutive no p-1,p,p+1 then
3(p+1)^2 = p-1+p +68
3p^2 +4p -64=0
thus p=4 by solving this
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