numbersThe first thing to do is to write the exact relations from the question, as follows:
[tex]\begin{gathered} x+y=42 \\ x\times y=432 \end{gathered}[/tex]Where X and Y stand for the unknown integer numbers.
Now, we will isolate Y in the first equation and substitute in the second one. This way, we will be able to find the value of X. From this strategy, we perform the calculation that follows:
[tex]\begin{gathered} y=42-x\to x\times(42-x)=432 \\ 42x-x^2=432\to0=x^2-42x+432 \\ x^2-42x+432=0 \end{gathered}[/tex]Now, it is important to remember the Bhaskara relation. But first, let's remember that any quadratic equation attends to the following generic form:
[tex]y=ax^2+bx+c[/tex]And we use the Bhaskara relation to find the values of X where Y is 0. In the present question, the constants are the following:
[tex]\begin{gathered} a=1 \\ b=-42 \\ c=432 \end{gathered}[/tex]And the Bhaskara relation is:
[tex]x_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]Now, we will substitute the values and perform the calculation.
[tex]\begin{gathered} x_{1,2}=\frac{-(-42)\pm\sqrt[]{(-42)^2-4\times1\times432}}{2\times1} \\ x_{1,2}=\frac{-(-42)\pm\sqrt[]{1,764-1,728}}{2}=\frac{42\pm\sqrt[]{36}}{2}=\frac{42\pm6}{2} \\ x_1=\frac{42+6}{2}=\frac{48}{2}=24_{} \\ x_2=\frac{42-6}{2}=\frac{36}{2}=18 \end{gathered}[/tex]As you can see, both numbers, 24 and 18, if summed will result in the number 42. For this reason, we found here, not only the value of X but also the value of Y. Because there is no distinction between X and Y, you say that:
