2.10-bo'lim : Radikallar bilan tenglamalar
2. Quyidagi tenglamani yeching.
\[\sqrt {33 - 2x} = x + 1\]Yechim
Bu erda birinchi qadam olish uchun ikkala tomonni kvadratga solishdir,
\[\begin{align*}{\left( {\sqrt {33 - 2x} } \right)^2} & = {\left( {x + 1} \right)^2}\\ 33 - 2x & = {x^2} + 2x + 1\\ {x^2} + 4x - 32 & = 0\end{align*}\]2-qadam
Bu shunchaki kvadrat tenglama va biz uni qanday hal qilishni bilamiz, shuning uchun buni qilaylik.
\[\left( {x + 8} \right)\left( {x - 4} \right) = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}x = - 8,\,\,\,\,\,x = 4\]Ko'rsatilgandek, biz birinchi qadamdan olingan kvadratikning ikkita echimiga egamiz.
3-qadam
Biz bu muammoni hal qilmadik. Eslatmalardan eslaylikki, biz bu erda qo'llagan yechim jarayoni ba'zan asl tenglamaga yechim bo'lmagan qiymatlarni kiritishning baxtsiz yon ta'siriga ega.
Shunday qilib, buni tugatish uchun biz asl tenglamaning oldingi bosqichidagi ikkala potentsial echimni tekshirishimiz kerak (eslang, asl tenglamadagi potentsial echimlarni tekshirish muhim).
\[x = - 8:\hspace{0.25in}\sqrt {33 - 2\left( { - 8} \right)} \mathop = \limits^? - 8 + 1\hspace{0.25in} \to \hspace{0.25in}\sqrt {49} \mathop = \limits^? - 7\hspace{0.25in} \to \hspace{0.25in}7 \ne - 7\hspace{0.25in}{\mbox{NOT}}{\mbox{ OK}}\] \[x = 4:\hspace{0.25in}\sqrt {33 - 2\left( 4 \right)} \mathop = \limits^? 4 + 1\hspace{0.25in} \to \hspace{0.25in}\sqrt {25} \mathop = \limits^? 5\hspace{0.25in} \to \hspace{0.25in}5 = 5\hspace{0.25in}{\mbox{OK}}\]Potentsial yechimlardan faqat bittasi ishlaydi va shuning uchun asl tenglama bitta yechimga ega: \(\require{bbox} \bbox[2pt,border:1px solid black]{{x = 4}}\).