2.9-bo'lim : Shaklda kvadratga tushiriladigan tenglamalar
2. Quyidagi tenglamani yeching.
\[{x^{ - 4}} - 7{x^{ - 2}} - 18 = 0\]
Maslahat: Birinchi ikkita hadning ko'rsatkichlariga qarashni unutmang va uni "normal" kvadrat tenglamaga aylantiradigan almashtirishni topishga harakat qiling.
Yechim
Birinchidan, \( - 4 = 2\left( { - 2} \right)\) va shuning uchun tenglamani kvadrat tenglamaga kamaytirish uchun quyidagi almashtirishdan foydalanishimiz mumkinligiga e'tibor bering.
\[u = {x^{ - 2}}\hspace{0.25in}\hspace{0.25in}{u^2} = {\left( {{x^{ - 2}}} \right)^2} = {x^{ - 4}}\]2-qadam
Ushbu almashtirishdan foydalanib, tenglama quyidagicha bo'ladi:
\[\begin{align*}{u^2} - 7u - 18 & = 0\\ \left( {u - 9} \right)\left( {u + 2} \right) & = 0\end{align*}\]Bu tenglamaning yechimi \(u = - 2\) va \(u = 9\) ekanligini osongina ko'rishimiz mumkin.
3-qadam
Endi biz qilishimiz kerak bo'lgan yagona narsa, dastlabki tenglamaning echimini aniqlash uchun birinchi bosqichdan boshlab almashtirishimizdan foydalanishdir.
\[u = - 2:\hspace{0.25in}{x^{ - 2}} = \frac{1}{{{x^2}}} = - 2\hspace{0.25in} \Rightarrow \hspace{0.25in}{x^2} = - \frac{1}{2}\hspace{0.25in} \Rightarrow \hspace{0.25in}x = \pm \sqrt { - \frac{1}{2}} = \pm \frac{1}{{\sqrt 2 }}i\] \[u = 9:\hspace{0.25in}{x^{ - 2}} = \frac{1}{{{x^2}}} = 9\hspace{0.25in} \Rightarrow \hspace{0.25in}{x^2} = \frac{1}{9}\hspace{0.25in} \Rightarrow \hspace{0.25in}x = \pm \sqrt {\frac{1}{9}} = \pm \frac{1}{3}\]Shuning uchun dastlabki tenglamaning to'rtta yechimi: \[\require{bbox} \bbox[2pt,border:1px solid black]{{x = \pm \frac{1}{{\sqrt 2 }}i\,\,{\mbox{and }}x = \pm \frac{1}{3}}}\].