(x+100)^2<100100000 eching

$$( x + 100 ) ^ { 2 } < 100100000$$

x uchun yechish

$$x\in \left(-100\sqrt{10010}-100,100\sqrt{10010}-100\right)$$

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.

$$\left(x+100\right)^{2}=0$$

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 200 ni va c uchun -100090000 ni ayiring.

$$x=\frac{-200±\sqrt{200^{2}-4\times 1\left(-100090000\right)}}{2}$$

Hisoblarni amalga oshiring.

$$x=\frac{-200±200\sqrt{10010}}{2}$$

x=\frac{-200±200\sqrt{10010}}{2} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$x=100\sqrt{10010}-100 x=-100\sqrt{10010}-100$$

Yechimlardan foydalanib tengsizlikni qaytadan yozing.

$$\left(x-\left(100\sqrt{10010}-100\right)\right)\left(x-\left(-100\sqrt{10010}-100\right)\right)<0$$

Koʻpaytma manfiy boʻlishi uchun x-\left(100\sqrt{10010}-100\right) va x-\left(-100\sqrt{10010}-100\right) qarama-qarshi belgilar boʻlishi kerak. x-\left(100\sqrt{10010}-100\right) musbat, x-\left(-100\sqrt{10010}-100\right) manfiy boʻlganda, yechimni toping.

$$x-\left(100\sqrt{10010}-100\right)>0 x-\left(-100\sqrt{10010}-100\right)<0$$

Bu har qanday x uchun xato.

$$x\in \emptyset$$

x-\left(-100\sqrt{10010}-100\right) musbat, x-\left(100\sqrt{10010}-100\right) manfiy boʻlganda, yechimni toping.

$$x-\left(-100\sqrt{10010}-100\right)>0 x-\left(100\sqrt{10010}-100\right)<0$$

Ikkala tengsizlikning mos yechimi – x\in \left(-100\sqrt{10010}-100,100\sqrt{10010}-100\right).

$$x\in \left(-100\sqrt{10010}-100,100\sqrt{10010}-100\right)$$

Oxirgi yechim olingan yechimlarning birlashmasidir.

$$x\in \left(-100\sqrt{10010}-100,100\sqrt{10010}-100\right)$$