Determine two consecutive even positive integers,the sum of whose squares is100
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ANSWER:--
What are the two consecutive even numbers the sum of whose squares is 100?
There are two possible sets of consecutive numbers that satisfy this question. While 6 and 8 will work, so will -8 and -6.
There are two possible sets of consecutive numbers that satisfy this question. While 6 and 8 will work, so will -8 and -6.Allow the two numbers to be x, and x+2. So:----
Allow the two numbers to be x, and x+2. So:
Allow the two numbers to be x, and x+2. So:x²+(x+2)²=100
Allow the two numbers to be x, and x+2. So:x²+(x+2)²=100x²+x²+4x+4–4=100–4
Allow the two numbers to be x, and x+2. So:x²+(x+2)²=100x²+x²+4x+4–4=100–42x²/2+4x/2=96/2
Allow the two numbers to be x, and x+2. So:x²+(x+2)²=100x²+x²+4x+4–4=100–42x²/2+4x/2=96/2x²+2x-48=48–48
Allow the two numbers to be x, and x+2. So:x²+(x+2)²=100x²+x²+4x+4–4=100–42x²/2+4x/2=96/2x²+2x-48=48–48x²+2x-48=0
Allow the two numbers to be x, and x+2. So:x²+(x+2)²=100x²+x²+4x+4–4=100–42x²/2+4x/2=96/2x²+2x-48=48–48x²+2x-48=0(x-6)(x+8)=0
Allow the two numbers to be x, and x+2. So:x²+(x+2)²=100x²+x²+4x+4–4=100–42x²/2+4x/2=96/2x²+2x-48=48–48x²+2x-48=0(x-6)(x+8)=0(6–6)(6+8)=0*8=0
Allow the two numbers to be x, and x+2. So:x²+(x+2)²=100x²+x²+4x+4–4=100–42x²/2+4x/2=96/2x²+2x-48=48–48x²+2x-48=0(x-6)(x+8)=0(6–6)(6+8)=0*8=0(-8–6)(-8+8)=-48*0=0
Allow the two numbers to be x, and x+2. So:x²+(x+2)²=100x²+x²+4x+4–4=100–42x²/2+4x/2=96/2x²+2x-48=48–48x²+2x-48=0(x-6)(x+8)=0(6–6)(6+8)=0*8=0(-8–6)(-8+8)=-48*0=0So this means that the value of x can be 6, or -8 and get valid results for either.