two consecutive positive even integers, the sum of whose squares is 340, we need to find the integers. give the quadratic equation formed by the problem (step by step)
Answers
Answered by
93
let x and x+2 two positive even numbers
sum of the squares of these numbers=340
x²+(x+2)²=340
x²+x²+4x+4 -340=0
2x²+4x-336=0
divide each term with 2
x²+2x-168=0
x²-14x+12x-14*12=0
x(x-14)+12(x-14)=0
(x-14)(x+12)=0
x-14 =0 or x+12=0
x=14 or x= -12
therefore
required numbers are x,x+2
if x=14 then 14 and 16
if x=-12 then -12 and -10
sum of the squares of these numbers=340
x²+(x+2)²=340
x²+x²+4x+4 -340=0
2x²+4x-336=0
divide each term with 2
x²+2x-168=0
x²-14x+12x-14*12=0
x(x-14)+12(x-14)=0
(x-14)(x+12)=0
x-14 =0 or x+12=0
x=14 or x= -12
therefore
required numbers are x,x+2
if x=14 then 14 and 16
if x=-12 then -12 and -10
Answered by
20
Answer:
let the numbers be x^2, (x+2)^2
according to the question,
x^2+(x+2)^2=340
by the formula of a^2+2ab+b^2
x^2+x^2+4x+4=340
2x^2+4x-336=0
÷ by 2
x^2+2x-168=0
x^2-14x+12x-168=0
x(x-14)+12(x-14)=0
(x+12)(x-14)
x=-12, 14
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