x^2-x-1>0 eching

$${ x }^{ 2 } -x-1 > 0$$

x uchun yechish

$$x\in \left(-\infty,\frac{1-\sqrt{5}}{2}\right)\cup \left(\frac{\sqrt{5}+1}{2},\infty\right)$$

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.

$$x^{2}-x-1=0$$

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

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

Hisoblarni amalga oshiring.

$$x=\frac{1±\sqrt{5}}{2}$$

x=\frac{1±\sqrt{5}}{2} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$x=\frac{\sqrt{5}+1}{2} x=\frac{1-\sqrt{5}}{2}$$

Yechimlardan foydalanib tengsizlikni qaytadan yozing.

$$\left(x-\frac{\sqrt{5}+1}{2}\right)\left(x-\frac{1-\sqrt{5}}{2}\right)>0$$

Koʻpaytma musbat boʻlishi uchun x-\frac{\sqrt{5}+1}{2} va x-\frac{1-\sqrt{5}}{2} ikkalasi yo manfiy, yo musbat boʻlishi kerak. x-\frac{\sqrt{5}+1}{2} va x-\frac{1-\sqrt{5}}{2} ikkalasi manfiy boʻlganda, yechimini toping.

$$x-\frac{\sqrt{5}+1}{2}<0 x-\frac{1-\sqrt{5}}{2}<0$$

Ikkala tengsizlikning mos yechimi – x<\frac{1-\sqrt{5}}{2}.

$$x<\frac{1-\sqrt{5}}{2}$$

x-\frac{\sqrt{5}+1}{2} va x-\frac{1-\sqrt{5}}{2} ikkalasi musbat boʻlganda, yechimini toping.

$$x-\frac{1-\sqrt{5}}{2}>0 x-\frac{\sqrt{5}+1}{2}>0$$

Ikkala tengsizlikning mos yechimi – x>\frac{\sqrt{5}+1}{2}.

$$x>\frac{\sqrt{5}+1}{2}$$

Oxirgi yechim olingan yechimlarning birlashmasidir.

$$x<\frac{1-\sqrt{5}}{2}\text{; }x>\frac{\sqrt{5}+1}{2}$$