53x^2+5x-12<0 eching

$$53 x ^ { 2 } + 5 x - 12 < 0$$

x uchun yechish

$$x\in \left(\frac{-\sqrt{2569}-5}{106},\frac{\sqrt{2569}-5}{106}\right)$$

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.

$$53x^{2}+5x-12=0$$

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni bu formula bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat tenglamada a uchun 53 ni, b uchun 5 ni va c uchun -12 ni ayiring.

$$x=\frac{-5±\sqrt{5^{2}-4\times 53\left(-12\right)}}{2\times 53}$$

Hisoblarni amalga oshiring.

$$x=\frac{-5±\sqrt{2569}}{106}$$

x=\frac{-5±\sqrt{2569}}{106} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$x=\frac{\sqrt{2569}-5}{106} x=\frac{-\sqrt{2569}-5}{106}$$

Yechimlardan foydalanib tengsizlikni qaytadan yozing.

$$53\left(x-\frac{\sqrt{2569}-5}{106}\right)\left(x-\frac{-\sqrt{2569}-5}{106}\right)<0$$

Koʻpaytma manfiy boʻlishi uchun x-\frac{\sqrt{2569}-5}{106} va x-\frac{-\sqrt{2569}-5}{106} qarama-qarshi belgilar boʻlishi kerak. x-\frac{\sqrt{2569}-5}{106} musbat, x-\frac{-\sqrt{2569}-5}{106} manfiy boʻlganda, yechimni toping.

$$x-\frac{\sqrt{2569}-5}{106}>0 x-\frac{-\sqrt{2569}-5}{106}<0$$

Bu har qanday x uchun xato.

$$x\in \emptyset$$

x-\frac{-\sqrt{2569}-5}{106} musbat, x-\frac{\sqrt{2569}-5}{106} manfiy boʻlganda, yechimni toping.

$$x-\frac{-\sqrt{2569}-5}{106}>0 x-\frac{\sqrt{2569}-5}{106}<0$$

Ikkala tengsizlikning mos yechimi – x\in \left(\frac{-\sqrt{2569}-5}{106},\frac{\sqrt{2569}-5}{106}\right).

$$x\in \left(\frac{-\sqrt{2569}-5}{106},\frac{\sqrt{2569}-5}{106}\right)$$

Oxirgi yechim olingan yechimlarning birlashmasidir.

$$x\in \left(\frac{-\sqrt{2569}-5}{106},\frac{\sqrt{2569}-5}{106}\right)$$