(x+3)^2<2 eching

$$( x + 3 ) ^ { 2 } < 2$$

x uchun yechish

$$x\in \left(-\sqrt{2}-3,\sqrt{2}-3\right)$$

Yechim qadamlari

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.

$$\left(x+3\right)^{2}=0$$

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 6 ni va c uchun 7 ni ayiring.

$$x=\frac{-6±\sqrt{6^{2}-4\times 1\times 7}}{2}$$

Hisoblarni amalga oshiring.

$$x=\frac{-6±2\sqrt{2}}{2}$$

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

$$x=\sqrt{2}-3 x=-\sqrt{2}-3$$

Yechimlardan foydalanib tengsizlikni qaytadan yozing.

$$\left(x-\left(\sqrt{2}-3\right)\right)\left(x-\left(-\sqrt{2}-3\right)\right)<0$$

Koʻpaytma manfiy boʻlishi uchun x-\left(\sqrt{2}-3\right) va x-\left(-\sqrt{2}-3\right) qarama-qarshi belgilar boʻlishi kerak. x-\left(\sqrt{2}-3\right) musbat, x-\left(-\sqrt{2}-3\right) manfiy boʻlganda, yechimni toping.

$$x-\left(\sqrt{2}-3\right)>0 x-\left(-\sqrt{2}-3\right)<0$$

Bu har qanday x uchun xato.

$$x\in \emptyset$$

x-\left(-\sqrt{2}-3\right) musbat, x-\left(\sqrt{2}-3\right) manfiy boʻlganda, yechimni toping.

$$x-\left(-\sqrt{2}-3\right)>0 x-\left(\sqrt{2}-3\right)<0$$

Ikkala tengsizlikning mos yechimi – x\in \left(-\left(\sqrt{2}+3\right),\sqrt{2}-3\right).

$$x\in \left(-\left(\sqrt{2}+3\right),\sqrt{2}-3\right)$$

Oxirgi yechim olingan yechimlarning birlashmasidir.

$$x\in \left(-\sqrt{2}-3,\sqrt{2}-3\right)$$