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the product of two numbers is 8 and their difference is 6. find the numbers.

Sagot :

[tex]the \ first\ number : \ x \\the \ second \ number : \ y \\\\ \begin{cases}x\cdot y =8 \\ x-y= 6 \end{cases} \\\\ \begin{cases}x\cdot y =8 \\ x = 6 + y \end{cases} \\\\ \begin{cases}x\cdot y(6+y)=8 \\ x = 6 + y \end{cases}[/tex]

[tex]y(6+y) =8 \\6y+y^2=8\\ y^2+6y-8=0\\a=1,\ \ b=6, \ \ c=-8\\\\y_{1}=\frac{-b-\sqrt{b^{2}-4ac}}{2a} =\frac{-6-\sqrt{6^2-4\cdot 1\cdot (-8)}}{2}= \frac{-6-\sqrt{36+32}}{2}=\frac{-6-\sqrt{68}}{2}=\\\\=\frac{-6-\sqrt{4\cdot 17}}{2}=\frac{-6-2\sqrt{17}}{2}=\frac{-2(3+\sqrt{17})}{2}=-(3+\sqrt{17})[/tex]

[tex]y_{2}=\frac{-b+\sqrt{b^{2}-4ac}}{2a} =\frac{-6+\sqrt{6^2-4\cdot 1\cdot (-8)}}{2}=\frac{2(-3+\sqrt{17})}{2}= \sqrt{17}-3 \\\\\begin{cases}x\cdot y =8 \\ y=-(3+\sqrt{17} )\end{cases} \ \ \ or \ \ \ \begin{cases}x\cdot y =8 \\ y= \sqrt{17}-3 \end{cases} \\\\ \begin{cases}x\cdot -( 3+\sqrt{17}) =8 \\ y=-(3+\sqrt{17}) \end{cases} \ \ \ or \ \ \ \begin{cases}x\cdot ( \sqrt{17}-3) =8 \\ y=\sqrt{17}-3 \end{cases}[/tex]

[tex]\begin{cases}x =\frac{8}{-(3+\sqrt{17})} \\ y=-3-\sqrt{17} \end{cases} \ \ \ or \ \ \ \begin{cases}x= \frac{8}{\sqrt{17}-3 } \\ y=\sqrt{17}-3 \end{cases}\\\\ \begin{cases}x =\frac{8}{-(3+\sqrt{17})} \\ y=-(3+\sqrt{17}) \end{cases} \ \ \ or \ \ \ \begin{cases}x= \frac{8}{\sqrt{17}-3 } \\ y=\sqrt{17}-3 \end{cases}[/tex]

[tex]\begin{cases}x =\frac{8}{-(3+\sqrt{17})}*\frac{3-\sqrt{17}}{3-\sqrt{17}} \\ y=-(3+\sqrt{17}) \end{cases} \ \ \ or \ \ \ \begin{cases}x= \frac{8}{\sqrt{17}-3 }*\frac{\sqrt{17}+3}{\sqrt{17}+3} \\ y=\sqrt{17}-3 \end{cases}[/tex]

[tex]\begin{cases}x = \frac{8(3-\sqrt{17})}{-(9-17)} \\ y=-(3+\sqrt{17}) \end{cases} \ \ \ or \ \ \ \begin{cases}x= \frac{8(\sqrt{17}+3)}{17-9} \\ y=\sqrt{17}- 3\end{cases}\\\\\begin{cases}x = \frac{8(3-\sqrt{17}) }{8 } \\ y=-(3+\sqrt{17}) \end{cases} \ \ \ or \ \ \ \begin{cases}x= \frac{8(\sqrt{17}+3) }{8} \\ y=\sqrt{17}- 3\end{cases}\\\\\begin{cases}x = \frac{8(3-\sqrt{17}) }{8 } \\ y=-(3+\sqrt{17}) \end{cases} \ \ \ or \ \ \ \begin{cases}x= \frac{8(\sqrt{17}+3)}{8} \\ y=\sqrt{17}- 3\end{cases}[/tex]

[tex]\begin{cases}x = 3-\sqrt{17} \\ y=-(3+\sqrt{17}) \end{cases} \ \ \ or \ \ \ \begin{cases}x= \sqrt{17}+3 \\ y=\sqrt{17}- 3\end{cases}[/tex]