(x-3)^2<2 eching

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

x uchun yechish

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

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.

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

Hisoblarni amalga oshiring.

$$x=\frac{6±2\sqrt{2}}{2}$$

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

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

Yechimlardan foydalanib tengsizlikni qaytadan yozing.

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

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

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

Bu har qanday x uchun xato.

$$x\in \emptyset$$

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

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

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

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

Oxirgi yechim olingan yechimlarning birlashmasidir.

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