Question

Solve application problems using quadratic equations. 1. A positive real number is 4 less than another. When 8 times the larger is added to the square of the smaller, the result is 96. Find the numbers.

Answer 1

Hello, let's  note a the positive real number.

A positive real number is 4 less than another.

So, the other number is a - 4

When 8 times the larger is added to the square of the smaller, the result is 96.

So, we need to solve the following equation.

$8a+(a-4)^2=96$

Let's do it!

$8a+(a-4)^2=96\\\\8a+a^2-8a+16=96\\\\a^2=96-16=80=4^2*5\\\\a=\pm4\sqrt{5}$

As a is positive, there is only one solution which is.

$\Large \boxed{\sf \bf \ \ a=4\sqrt{5} \ \ }$

Thank you.