2.6-bo'lim : Kvadrat tenglamalar - II qism
8. Kvadratni to‘ldirib, quyidagi kvadrat tenglamani yeching.
\[2{x^2} + 5x + 3 = 0\]Yechim
Birinchidan, barcha o'zgaruvchilar bir tomonda, \({x^2}\) koeffitsienti bir, raqam esa boshqa tomonda bo'lgan tenglamani shaklga keltiramiz. Shunday qilib,
\[\begin{align*}{x^2} + \frac{5}{2}x + \frac{3}{2} & = 0\\ {x^2} + \frac{5}{2}x & = - \frac{3}{2}\end{align*}\]2-qadam
Endi biz tenglamaning chap tomonidagi ifodadagi kvadratni to'ldirishimiz mumkin.
Buni qilishimiz kerak bo'lgan raqam,
\[{\left( {\frac{{\frac{5}{2}}}{2}} \right)^2} = {\left( {\frac{5}{4}} \right)^2} = \frac{{25}}{{16}}\]3-qadam
Shu nuqtada biz bu erda tenglama borligini va tenglamaning bir tomoniga nima qilsak, ikkinchisini ham qilishimiz kerakligini esga olishimiz kerak. Boshqacha qilib aytadigan bo'lsak, 1-bosqichdagi tenglamaning ikkala tomoniga oldingi bosqichdagi raqamni qo'shishni unutmang.
\[\require{color}\begin{align*}{x^2} + \frac{5}{2}x \,{\color{Red} + \frac{{25}}{{16}}} & = - \frac{3}{2} \,{\color{Red} + \frac{{25}}{{16}}}\\ {\left( {x + \frac{5}{4}} \right)^2} & = \frac{1}{{16}}\end{align*}\]4-qadam
Endi tenglamani yechish uchun faqat oldingi bosqichdagi tenglamada Kvadrat ildiz xususiyatidan foydalanish kerak. Shunday qilib,
\[\begin{align*}x + \frac{5}{4} & = \pm \sqrt {\frac{1}{{16}}} = \pm \frac{1}{4}\\ x & = - \frac{5}{4} \pm \frac{1}{4}\hspace{0.25in} \Rightarrow \hspace{0.25in}x = - \frac{5}{4} - \frac{1}{4} = - \frac{3}{2},\,\,\,\,\,\,\,\,x = - \frac{5}{4} + \frac{1}{4} = - 1\end{align*}\]Bu tenglamaning ikkita yechimi quyidagicha bo'ladi: \[\require{bbox} \bbox[2pt,border:1px solid black]{{x = - \frac{3}{2}\,\,{\mbox{and }}x = - 1}}\].