Question 1163436
<br>
A positive real number is 4 less than another.<br>
Let the two numbers be x and x-4.<br>
When 8 times the larger is added to the square of the smaller, the result is 96.<br>
{{{(x-4)^2+8x = 96}}}<br>
Find the numbers.<br>
{{{x^2-8x+16+8x = 96}}}
{{{x^2+16 = 96}}}
{{{x^2 = 80}}}
{{{x = sqrt(80) = 4*sqrt(5)}}}<br>
Note the problem says x is positive, so we ignore the negative value.<br>
ANSWERS: The two numbers are {{{4*sqrt(5)}}} and {{{4*sqrt(5)-4}}}<br>
CHECK:
{{{(4*sqrt(5)-4)^2+8(4*sqrt(5)) = 80-32*sqrt(5)+16+32*sqrt(5) = 96}}}<br>