Question 92505
Translate the word problem:

{{{x^2+(x+1)^2=41}}} "two consecutive positive integers such that the sum of their squares is 41"


{{{x^2+x^2+2x+1=41}}} Foil


{{{x^2+x^2+2x+1-41=0}}} Subtract 41 from both sides


{{{2x^2+2x-40=0}}} Combine like terms



Let's use the quadratic formula to solve for x:



Starting with the general quadratic


{{{ax^2+bx+c=0}}}


the general solution using the quadratic equation is:


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a)}}}


So lets solve {{{2*x^2+2*x-40=0}}} ( notice {{{a=2}}}, {{{b=2}}}, and {{{c=-40}}})


{{{x = (-2 +- sqrt( (2)^2-4*2*-40 ))/(2*2)}}} Plug in a=2, b=2, and c=-40




{{{x = (-2 +- sqrt( 4-4*2*-40 ))/(2*2)}}} Square 2 to get 4  




{{{x = (-2 +- sqrt( 4+320 ))/(2*2)}}} Multiply {{{-4*-40*2}}} to get {{{320}}}




{{{x = (-2 +- sqrt( 324 ))/(2*2)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (-2 +- 18)/(2*2)}}} Simplify the square root (note: If you need help with simplifying the square root, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)




{{{x = (-2 +- 18)/4}}} Multiply 2 and 2 to get 4


So now the expression breaks down into two parts


{{{x = (-2 + 18)/4}}} or {{{x = (-2 - 18)/4}}}


Lets look at the first part:


{{{x=(-2 + 18)/4}}}


{{{x=16/4}}} Add the terms in the numerator

{{{x=4}}} Divide


So one answer is

{{{x=4}}}




Now lets look at the second part:


{{{x=(-2 - 18)/4}}}


{{{x=-20/4}}} Subtract the terms in the numerator

{{{x=-5}}} Divide


So another answer is

{{{x=-5}}}


So our possible solutions are:

{{{x=4}}} or {{{x=-5}}}


Since we only care about the positive numbers, our only solution is {{{x=4}}}



So if x=4 then...


{{{4+1=5}}} the 2nd number is 5


So the pair of numbers is 4 and 5



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Check:


{{{4^2+5^2=41}}} plug in the pair of numbers



{{{16+25=41}}} Square each term


{{{41=41}}} Combine like terms. Since the equation is true, our answer is verified.