12x^2-12^2x+9>0 eching

$$12 { x }^{ 2 } - { 12 }^{ 2 } x+9 > 0$$

x uchun yechish

$$x\in \left(-\infty,-\frac{\sqrt{141}}{2}+6\right)\cup \left(\frac{\sqrt{141}}{2}+6,\infty\right)$$

Kvadrat tenglamalaridan foydalanish qoidalari

2 daraja ko‘rsatkichini 12 ga hisoblang va 144 ni qiymatni oling.

$$12x^{2}-144x+9>0$$

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.

$$12x^{2}-144x+9=0$$

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni bu formula bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat tenglamada a uchun 12 ni, b uchun -144 ni va c uchun 9 ni ayiring.

$$x=\frac{-\left(-144\right)±\sqrt{\left(-144\right)^{2}-4\times 12\times 9}}{2\times 12}$$

Hisoblarni amalga oshiring.

$$x=\frac{144±12\sqrt{141}}{24}$$

x=\frac{144±12\sqrt{141}}{24} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$x=\frac{\sqrt{141}}{2}+6 x=-\frac{\sqrt{141}}{2}+6$$

Yechimlardan foydalanib tengsizlikni qaytadan yozing.

$$12\left(x-\left(\frac{\sqrt{141}}{2}+6\right)\right)\left(x-\left(-\frac{\sqrt{141}}{2}+6\right)\right)>0$$

Koʻpaytma musbat boʻlishi uchun x-\left(\frac{\sqrt{141}}{2}+6\right) va x-\left(-\frac{\sqrt{141}}{2}+6\right) ikkalasi yo manfiy, yo musbat boʻlishi kerak. x-\left(\frac{\sqrt{141}}{2}+6\right) va x-\left(-\frac{\sqrt{141}}{2}+6\right) ikkalasi manfiy boʻlganda, yechimini toping.

$$x-\left(\frac{\sqrt{141}}{2}+6\right)<0 x-\left(-\frac{\sqrt{141}}{2}+6\right)<0$$

Ikkala tengsizlikning mos yechimi – x<-\frac{\sqrt{141}}{2}+6.

$$x<-\frac{\sqrt{141}}{2}+6$$

x-\left(\frac{\sqrt{141}}{2}+6\right) va x-\left(-\frac{\sqrt{141}}{2}+6\right) ikkalasi musbat boʻlganda, yechimini toping.

$$x-\left(-\frac{\sqrt{141}}{2}+6\right)>0 x-\left(\frac{\sqrt{141}}{2}+6\right)>0$$

Ikkala tengsizlikning mos yechimi – x>\frac{\sqrt{141}}{2}+6.

$$x>\frac{\sqrt{141}}{2}+6$$

Oxirgi yechim olingan yechimlarning birlashmasidir.

$$x<-\frac{\sqrt{141}}{2}+6\text{; }x>\frac{\sqrt{141}}{2}+6$$