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Sagot :
Let X be one number and Y be other number. Then, we can write
[tex]X\cdot Y=160[/tex]and
[tex]X-Y=-4[/tex]By moving -Y to the right hand side, we get
[tex]X=Y-4[/tex]By substituting the last expression into the first one, we get
[tex](Y-4)\cdot Y=160[/tex]which gives
[tex]\begin{gathered} Y^2-4Y=160 \\ Y^2-4Y-160=0 \end{gathered}[/tex]So, we have a quadratic function, which we can solve by means of the quadractic formula:
[tex]Y=\frac{-(-4)\pm\sqrt[]{(-4)^2-4(1)(-160})}{2}[/tex]which gives
[tex]\begin{gathered} Y=\frac{4\pm\sqrt[]{16+640}}{2} \\ Y=\frac{4\pm25.6}{2} \end{gathered}[/tex]So, we have 2 solutions:
[tex]\begin{gathered} Y=\frac{29.6}{2}=14.8 \\ \text{and} \\ Y=\frac{-21.6}{2}=-10.8 \end{gathered}[/tex]By substituting the fist solution into our first equation, we get
[tex]\begin{gathered} X\cdot(14.81)=160 \\ X=\frac{160}{14.8} \\ X=10.81 \end{gathered}[/tex]Now, by substituting the second solution into our first equation, we h ave
[tex]\begin{gathered} X\cdot(-10.8)=160 \\ X=\frac{160}{-10.8} \\ X=-14.81 \end{gathered}[/tex]Lets check one solution. For X=10.81 and Y=14.81 we have
[tex]\begin{gathered} (10.81)(14.8)\approx160 \\ \text{and} \\ 110.81-14.8\approx-4 \end{gathered}[/tex]Now, lets chech the second solution. For X=-14.81 and Y=-10.81 we have
[tex]\begin{gathered} (-10.81)(-14.8)\approx160 \\ \text{and} \\ -14.81+10.81\approx-4 \end{gathered}[/tex]They are correct !! So, one solution is X=10.81 and Y=14.8 and the other solution is X=-14.81 and Y=-10.81.
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