Question

Find two positive numbers whose difference is 30 and whose product is 2584.

Answer 1

Answer:

$38$ and $68$.

Step-by-step explanation:

Let $x$ be the smaller one of the two number.

$x$ must be a positive integer. The other number would be $(x + 30)$.

The question states that the product of the two numbers is $2584$. In other words:

$x\, (x + 30) = 2584$.

Rearrange this equation and solve for $x$:

$x^{2} + 30\, x - 2584 = 0$.

The first root of this quadratic equation would be:

$\begin{aligned}x_{1} &= \frac{(-30) + \sqrt{30^{2} - 4 \times (-2584)}}{2} \\ &= \frac{(-30) + \sqrt{900 + 10336}}{2} \\ &= \frac{(-30) + \sqrt{11236}}{2} \\ &= \frac{(-30)}{2} + \sqrt{\frac{11236}{2^{2}}} \\ &= (-15) + \sqrt{2809} \\ &= (-15) + 53 \\ &= 38 \end{aligned}$.

Similarly, the second root of this quadratic equation would be:

$\begin{aligned}x_{1} &= \frac{(-30) - \sqrt{30^{2} - 4 \times (-2584)}}{2} \\ &= (-15) - 53 \\ &= -68\end{aligned}$.

Since the question requires that both numbers should be positive, $x > 0$. Therefore, only $x = 38$ is valid.

Hence, the two numbers would be $38$ and $(38 + 30)$, which is $68$.

Answer 2

9514 1404 393

Answer:

  38 and 68

Step-by-step explanation:

Let x represent the average of the two numbers. Then their product is ...

  (x -15)(x +15) = 2584

  x^2 = 2584 +225 = 2809 . . . . . . use relation (x -a)(x +a) = x^2 -a^2; add a^2

  x = √2809 = 53

Then the two numbers are ...

  53 +15 = 68

and

  68 -30 = 38

The two numbers are 38 and 68.

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Additional comment

Solving the problem in this way is equivalent to writing the quadratic equation for the smallest (or largest) number and solving that by completing the square. The "solution" x = -53 is not relevant in this problem.