Question 1178016

(1)ONE NUMBER EXCEEDS ANOTHER BY 5 THE SUM OF THEIR SQUARES IS 120,FIND THE NUMBERS. 

let first number be {{{x}}}
and second number {{{y}}}
 if second number exceeds first by {{{5}}}

so, we have {{{x=y-5}}}..........eq.1

the sum of their squares is {{{120}}}

{{{x^2+y^2=120}}}

{{{(y-5)^2+y^2=120}}}

{{{2 y^2 - 10 y + 25=120}}}

{{{2 y^2 - 10 y + 25-120=0}}}

{{{2 y^2 - 10 y -95=0}}}...........using quadratic formula we get

{{{y = 5/2 + sqrt(215)/2}}} or  {{{y=9.83143914930759}}}
or
{{{y = 5/2 - sqrt(215)/2}}} or   {{{y=-4.83143914930759}}}

then

{{{x=5/2 + sqrt(215)/2-5}}} or  {{{x=4.83143914930759}}}
or
{{{x=5/2 -sqrt(215)/2-5}}} or  {{{x=-9.83143914930759}}}

your numbers are:

{{{4.83143914930759}}} and {{{9.83143914930759}}}
or
{{{-4.83143914930759}}} and {{{-9.83143914930759}}}

check: for accuracy using all decimal places

{{{9.83143914930759^2+(-4.83143914930759)^2=120}}}
{{{120=120}}}


(2)TWO NUMBERS DIFFERS BY 6,THE DIFFERENCE OF THEIR SQUARES IS 120,FIND THE NUMBERS.

let first number be {{{x}}}
and second number {{{y}}}
 if second number differ by {{{6}}}

so, we have {{{y=x-6}}}..........eq.1

the difference of their squares is {{{120}}}

{{{x^2-y^2=120}}}

{{{x^2-(x-6)^2=120}}}
{{{x^2-(x^2-12x+36)=120}}}
{{{x^2-x^2+12x-36=120}}}
{{{12x-36=120}}}.........simplify, divide by {{{12}}}
{{{x-3=10}}}
{{{x=10+3}}}
{{{x=13}}}

then

{{{y=x-6}}}->{{{y=13-6}}}->{{{y=7}}}

your numbers are: {{{13}}} and {{{7}}}

check the difference of their squares:

{{{13^2-7^2=120}}}
{{{169-49=120}}}
{{{120=120}}}