11x^2-9x+1>0 eching

$$11 { x }^{ 2 } -9x+1 > 0$$

x uchun yechish

$$x\in \left(-\infty,\frac{9-\sqrt{37}}{22}\right)\cup \left(\frac{\sqrt{37}+9}{22},\infty\right)$$

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.

$$11x^{2}-9x+1=0$$

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni bu formula bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat tenglamada a uchun 11 ni, b uchun -9 ni va c uchun 1 ni ayiring.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 11\times 1}}{2\times 11}$$

Hisoblarni amalga oshiring.

$$x=\frac{9±\sqrt{37}}{22}$$

x=\frac{9±\sqrt{37}}{22} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$x=\frac{\sqrt{37}+9}{22} x=\frac{9-\sqrt{37}}{22}$$

Yechimlardan foydalanib tengsizlikni qaytadan yozing.

$$11\left(x-\frac{\sqrt{37}+9}{22}\right)\left(x-\frac{9-\sqrt{37}}{22}\right)>0$$

Koʻpaytma musbat boʻlishi uchun x-\frac{\sqrt{37}+9}{22} va x-\frac{9-\sqrt{37}}{22} ikkalasi yo manfiy, yo musbat boʻlishi kerak. x-\frac{\sqrt{37}+9}{22} va x-\frac{9-\sqrt{37}}{22} ikkalasi manfiy boʻlganda, yechimini toping.

$$x-\frac{\sqrt{37}+9}{22}<0 x-\frac{9-\sqrt{37}}{22}<0$$

Ikkala tengsizlikning mos yechimi – x<\frac{9-\sqrt{37}}{22}.

$$x<\frac{9-\sqrt{37}}{22}$$

x-\frac{\sqrt{37}+9}{22} va x-\frac{9-\sqrt{37}}{22} ikkalasi musbat boʻlganda, yechimini toping.

$$x-\frac{9-\sqrt{37}}{22}>0 x-\frac{\sqrt{37}+9}{22}>0$$

Ikkala tengsizlikning mos yechimi – x>\frac{\sqrt{37}+9}{22}.

$$x>\frac{\sqrt{37}+9}{22}$$

Oxirgi yechim olingan yechimlarning birlashmasidir.

$$x<\frac{9-\sqrt{37}}{22}\text{; }x>\frac{\sqrt{37}+9}{22}$$