27c^2-27c-22>0 eching

$$27 c ^ { 2 } - 27 c - 22 > 0$$

c uchun yechish

$$c\in \left(-\infty,-\frac{\sqrt{345}}{18}+\frac{1}{2}\right)\cup \left(\frac{\sqrt{345}}{18}+\frac{1}{2},\infty\right)$$

Kvadrat tenglamalaridan foydalanish qoidalari

Tengsizlikni yechish uchun chap tomon faktorini hisoblang. Kvadrat koʻp tenglama bu orqali hisoblanadi: ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), bu yerda x_{1} va x_{2} ax^{2}+bx+c=0 kvadrat tenglamaning yechimlari.

$$27c^{2}-27c-22=0$$

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni bu formula bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat tenglamada a uchun 27 ni, b uchun -27 ni va c uchun -22 ni ayiring.

$$c=\frac{-\left(-27\right)±\sqrt{\left(-27\right)^{2}-4\times 27\left(-22\right)}}{2\times 27}$$

Hisoblarni amalga oshiring.

$$c=\frac{27±3\sqrt{345}}{54}$$

c=\frac{27±3\sqrt{345}}{54} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$c=\frac{\sqrt{345}}{18}+\frac{1}{2} c=-\frac{\sqrt{345}}{18}+\frac{1}{2}$$

Yechimlardan foydalanib tengsizlikni qaytadan yozing.

$$27\left(c-\left(\frac{\sqrt{345}}{18}+\frac{1}{2}\right)\right)\left(c-\left(-\frac{\sqrt{345}}{18}+\frac{1}{2}\right)\right)>0$$

Koʻpaytma musbat boʻlishi uchun c-\left(\frac{\sqrt{345}}{18}+\frac{1}{2}\right) va c-\left(-\frac{\sqrt{345}}{18}+\frac{1}{2}\right) ikkalasi yo manfiy, yo musbat boʻlishi kerak. c-\left(\frac{\sqrt{345}}{18}+\frac{1}{2}\right) va c-\left(-\frac{\sqrt{345}}{18}+\frac{1}{2}\right) ikkalasi manfiy boʻlganda, yechimini toping.

$$c-\left(\frac{\sqrt{345}}{18}+\frac{1}{2}\right)<0 c-\left(-\frac{\sqrt{345}}{18}+\frac{1}{2}\right)<0$$

Ikkala tengsizlikning mos yechimi – c<-\frac{\sqrt{345}}{18}+\frac{1}{2}.

$$c<-\frac{\sqrt{345}}{18}+\frac{1}{2}$$

c-\left(\frac{\sqrt{345}}{18}+\frac{1}{2}\right) va c-\left(-\frac{\sqrt{345}}{18}+\frac{1}{2}\right) ikkalasi musbat boʻlganda, yechimini toping.

$$c-\left(-\frac{\sqrt{345}}{18}+\frac{1}{2}\right)>0 c-\left(\frac{\sqrt{345}}{18}+\frac{1}{2}\right)>0$$

Ikkala tengsizlikning mos yechimi – c>\frac{\sqrt{345}}{18}+\frac{1}{2}.

$$c>\frac{\sqrt{345}}{18}+\frac{1}{2}$$

Oxirgi yechim olingan yechimlarning birlashmasidir.

$$c<-\frac{\sqrt{345}}{18}+\frac{1}{2}\text{; }c>\frac{\sqrt{345}}{18}+\frac{1}{2}$$