x^2-2x-5<0 eching

$$x ^ { 2 } - 2 x - 5 < 0$$

x uchun yechish

$$x\in \left(1-\sqrt{6},\sqrt{6}+1\right)$$

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.

$$x^{2}-2x-5=0$$

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 -2 ni va c uchun -5 ni ayiring.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 1\left(-5\right)}}{2}$$

Hisoblarni amalga oshiring.

$$x=\frac{2±2\sqrt{6}}{2}$$

x=\frac{2±2\sqrt{6}}{2} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$x=\sqrt{6}+1 x=1-\sqrt{6}$$

Yechimlardan foydalanib tengsizlikni qaytadan yozing.

$$\left(x-\left(\sqrt{6}+1\right)\right)\left(x-\left(1-\sqrt{6}\right)\right)<0$$

Koʻpaytma manfiy boʻlishi uchun x-\left(\sqrt{6}+1\right) va x-\left(1-\sqrt{6}\right) qarama-qarshi belgilar boʻlishi kerak. x-\left(\sqrt{6}+1\right) musbat, x-\left(1-\sqrt{6}\right) manfiy boʻlganda, yechimni toping.

$$x-\left(\sqrt{6}+1\right)>0 x-\left(1-\sqrt{6}\right)<0$$

Bu har qanday x uchun xato.

$$x\in \emptyset$$

x-\left(1-\sqrt{6}\right) musbat, x-\left(\sqrt{6}+1\right) manfiy boʻlganda, yechimni toping.

$$x-\left(1-\sqrt{6}\right)>0 x-\left(\sqrt{6}+1\right)<0$$

Ikkala tengsizlikning mos yechimi – x\in \left(1-\sqrt{6},\sqrt{6}+1\right).

$$x\in \left(1-\sqrt{6},\sqrt{6}+1\right)$$

Oxirgi yechim olingan yechimlarning birlashmasidir.

$$x\in \left(1-\sqrt{6},\sqrt{6}+1\right)$$