Question

A positive real number is 3 less than another. When 6 × the larger is added to the square of the smaller, the result is 63. Find the numbers.

A) 6, 3
B) 5, 2
C) 4, 1
D) 7, 4

Answer 1

The two positive real numbers are found by first expressing the larger number as the smaller number plus 3 and then forming and solving a quadratic equation. The smaller number turns out to be 5, and when 3 is added, the larger number is 8.

Step-by-step explanation:

Let's denote the smaller real number as x and the larger real number as y. According to the problem, y is 3 more than x, so we can write it as y = x + 3. Now, when 6 times the larger is added to the square of the smaller, the result is 63. The equation for this scenario is 6y + x2 = 63.

Substituting y with (x + 3) in the equation, we get:

6(x + 3) + x2 = 63

Expanding this, we obtain:

6x + 18 + x2 = 63

Bringing all terms to one side to set the equation to zero, we have:

x2 + 6x + 18 - 63 = 0

Simplifying further:

x2 + 6x - 45 = 0

Factoring this quadratic equation, we find:

(x + 9)(x - 5) = 0

Setting each factor equal to zero gives us two potential solutions for x:

• x + 9 = 0 → x = -9 (not possible since the number is positive)
• x - 5 = 0 → x = 5 (this is a positive number)

Since only the positive value of x is acceptable, we then find y:

y = x + 3

y = 5 + 3

y = 8

Hence, the two numbers are 5 and 8.