62x^2+3x-1<0 eching

$$62 x ^ { 2 } + 3 x - 1 < 0$$

x uchun yechish

$$x\in \left(\frac{-\sqrt{257}-3}{124},\frac{\sqrt{257}-3}{124}\right)$$

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.

$$62x^{2}+3x-1=0$$

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni bu formula bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat tenglamada a uchun 62 ni, b uchun 3 ni va c uchun -1 ni ayiring.

$$x=\frac{-3±\sqrt{3^{2}-4\times 62\left(-1\right)}}{2\times 62}$$

Hisoblarni amalga oshiring.

$$x=\frac{-3±\sqrt{257}}{124}$$

x=\frac{-3±\sqrt{257}}{124} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$x=\frac{\sqrt{257}-3}{124} x=\frac{-\sqrt{257}-3}{124}$$

Yechimlardan foydalanib tengsizlikni qaytadan yozing.

$$62\left(x-\frac{\sqrt{257}-3}{124}\right)\left(x-\frac{-\sqrt{257}-3}{124}\right)<0$$

Koʻpaytma manfiy boʻlishi uchun x-\frac{\sqrt{257}-3}{124} va x-\frac{-\sqrt{257}-3}{124} qarama-qarshi belgilar boʻlishi kerak. x-\frac{\sqrt{257}-3}{124} musbat, x-\frac{-\sqrt{257}-3}{124} manfiy boʻlganda, yechimni toping.

$$x-\frac{\sqrt{257}-3}{124}>0 x-\frac{-\sqrt{257}-3}{124}<0$$

Bu har qanday x uchun xato.

$$x\in \emptyset$$

x-\frac{-\sqrt{257}-3}{124} musbat, x-\frac{\sqrt{257}-3}{124} manfiy boʻlganda, yechimni toping.

$$x-\frac{-\sqrt{257}-3}{124}>0 x-\frac{\sqrt{257}-3}{124}<0$$

Ikkala tengsizlikning mos yechimi – x\in \left(\frac{-\sqrt{257}-3}{124},\frac{\sqrt{257}-3}{124}\right).

$$x\in \left(\frac{-\sqrt{257}-3}{124},\frac{\sqrt{257}-3}{124}\right)$$

Oxirgi yechim olingan yechimlarning birlashmasidir.

$$x\in \left(\frac{-\sqrt{257}-3}{124},\frac{\sqrt{257}-3}{124}\right)$$