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.

Answer 2

The amount given by the sum of numbers can be presented in an equation

form from which the value of each number can be obtained.

The large number is 4·√5, the smaller number is 4·(√5 - 1)

Reason:

Let x represent the large number, we have;

The smaller number = x - 4

8 × x + (x - 4)² = 96

8·x + (x - 4)² = 96

Required:

The values of the numbers

Solution:

8·x + (x - 4)² = 8·x + x² - 8·x + 16 = 96

8·x - 8·x + x² + 16 = 96

x² + 16 = 96

x² = 96 - 16 = 80

x = ±√(80) = ±4·√5

Given that the numbers are positive, we have;

The large number, x = 4·√5

The smaller number = x - 4 = 4·√5 - 4

The smaller number = 4·√5 - 4 = 4·(√5 - 1)

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