91Ó°ÊÓ

If $a{x}^2+bx+c=0$, with $a≠0$, then $x=\frac{−bÂ\pm \sqrt{b^2−4ac}}{2a}$. Solving using factoring can be done by splitting the middle term of the equation and then making groups and equating them to zero.

Let $x$and $y$be two numbers. Then the system of equations formed as per the information given is:

role="math" localid="1647850417864" $$\begin{gathered} x+y=−2…\left(1\right) \\ x×y=−48…\left(2\right) \end{gathered}$$

Write equation (1) in terms of $x$.

role="math" localid="1647850479875" $$\begin{gathered} x+y=−\text{ â¶Ä‰}2 \\ x\text{ â¶Ä‰}+y−y=\text{ â¶Ä‰}−\text{ â¶Ä‰}2−y \\ x=−\text{ â¶Ä‰}2−y \end{gathered}$$

Now substitute the value of $x$into equation (2).

$$\begin{gathered} x×y\text{ â¶Ä‰}=\text{ â¶Ä‰}−48 \\ (−\text{ â¶Ä‰}2−y)\text{ â¶Ä‰}·\text{ â¶Ä‰}y\text{ â¶Ä‰}=\text{ â¶Ä‰}−48 \\ −\text{ â¶Ä‰}2y−y^2=−\text{ â¶Ä‰}48\text{ â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰}\left[\text{ Distribute â¶Ä‰}\right] \end{gathered}$$

Simplify the quadratic equation for $y$. by completing the square.

$−\text{ â¶Ä‰}2y+2y\text{ â¶Ä‰}−y^2+y^2\text{ â¶Ä‰}=−\text{ â¶Ä‰}48+2y\text{ â¶Ä‰}+y^2\text{ â¶Ä‰}\left[\text{ â¶Ä‰make â¶Ä‰one â¶Ä‰side â¶Ä‰equal â¶Ä‰to â¶Ä‰zero â¶Ä‰}\right]$

$$\begin{gathered} 0=\text{ â¶Ä‰}{y}^2\text{ }+2y−48 \\ 0=(y−6)(y+8)\text{ â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰}\left[\text{ â¶Ä‰â€‰Factor â¶Ä‰}\right] \end{gathered}$$

Use the zero product rule to further find the values of $y$.

$y−6=0\text{ â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰and â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰}y+8=0$

$$\begin{gathered} y=6\text{ â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰and â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰}y=−8 \\ x=−2−y \\ x=−2−6\text{ â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰and â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰}x=−2−(−8) \\ x=−8\text{ â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰and â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰}x=6 \end{gathered}$$

The set of numbers is: $(−8,\text{ â¶Ä‰}6)$or $(6,-8)$.