Question

two numbers whose sum is 22 and its product is equal to 104​

Answer 1

I think you meant: "Two numbers whose sum is 21". If we consider that the sum is 22, there would be no whole numbers as results.

If you meant "21", the resolution is here. If the number really is "22", the resolution is at the end of the answer.

Two numbers whose sum is 21 and its product is equal to 104​

Let the first number be "x" and the second number is "21 - x".

According to the question, its product is equal to 104​:

         x (21 - x) = 104

          21x - x² = 104

-x² + 21x - 104 = 0

Now, let's change the sign of the entire equation:

• x² - 21x + 104 = 0

This is a second degree equation. We will use the general formula of complete second degree equations to solve it:

$\large{\boxed{\mathsf{x=\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}}}}$

Let's identify the constants. The constants are the numbers that go in front of x squared, x and the term that does not carry x (in the formula, the constants are a, b and c).

x² - 21x + 104 = 0

• In this exercise, there is nothing in front of x squared, therefore a = 1.
• In front of x there is a -21, so b = -21. We must consider the "minus" sign.
• And the term that does not carry x is 104, so c = 104.

Now, let's apply the formula:

$\mathsf{x=\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}}$

$\mathsf{x=\dfrac{-(-21)\pm \sqrt{(-21)^{2}-4(1)(104)}}{2(1)}}$

$\mathsf{x=\dfrac{21\pm \sqrt{441-416}}{2}}$

$\mathsf{x=\dfrac{21\pm \sqrt{25}}{2}}$

$\mathsf{x=\dfrac{21\pm 5}{2}}$

At this point, we have to resolve the + sign on the one hand and the – sign on the other:

 $\mathsf{x_1 = \dfrac{21 + 5}{2}}$      and      $\mathsf{x_2 = \dfrac{21-5}{2}}}$

$\boxed{\mathsf{x_1 = 13}}$                        $\boxed{\mathsf{x_2 = 8}}$

If one number is 13, the other would be 8, or vice versa.

Therefore, the numbers are 13 and 8.

Two numbers whose sum is 22 and its product is equal to 104​

Let the first number be "x" and the second number is "22 - x".

According to the question, its product is equal to 104​:

         x (22 - x) = 104

          22x - x² = 104

-x² + 22x - 104 = 0

Now, let's change the sign of the entire equation:

• x² - 22x + 104 = 0

This is a second degree equation. Let's use the general formula of complete second degree equations to solve it:

$\large{\boxed{\mathsf{x=\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}}}}$

Let's identify the constants. The constants are the numbers that go in front of x squared, x and the term that does not carry x (in the formula, the constants are a, b and c).

x² - 22x + 104 = 0

• In this exercise, there is nothing in front of x squared, therefore a = 1.
• In front of x there is a -22, so b = -22. We must consider the "minus" sign.
• And the term that does not carry x is 104, so c = 104.

Now, let's apply the formula:

$\mathsf{x=\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}}$

$\mathsf{x=\dfrac{-(-22)\pm \sqrt{(-22)^{2}-4(1)(104)}}{2(1)}}$

$\mathsf{x=\dfrac{22\pm \sqrt{484-416}}{2}}$

$\mathsf{x=\dfrac{22\pm \sqrt{68}}{2}}$

$\mathrm{[Let's\ simplify\ the\ root:\ \sqrt{68} = \sqrt{2^{2} \cdot 17} = \sqrt{2^{2}}\cdot\sqrt{17}=\boxed{2\sqrt{17}}]}$

$\mathsf{x=\dfrac{22\pm 2\sqrt{17}}{2}}$

Let's simplify the fraction:

$\mathsf{x=\dfrac{22}{2}\pm \dfrac{2\sqrt{17}}{2}}$

$\boxed{\mathsf{x = 11 \pm \sqrt{17}}}}}$

At this point, we have to resolve the + sign on the one hand and the – sign on the other:

$\boxed{\mathsf{x_1 = 11 + \sqrt{17}}}$      and      $\boxed{\mathsf{x_2 = 11 - \sqrt{17}}}$

The numbers are 11 + √17 and 11 - √17.