2x^2-6>-3x eching

$$2 x ^ { 2 } - 6 > - 3 x$$

x uchun yechish

$$x\in \left(-\infty,\frac{-\sqrt{57}-3}{4}\right)\cup \left(\frac{\sqrt{57}-3}{4},\infty\right)$$

Kvadrat tenglamalaridan foydalanish qoidalari

3x ni ikki tarafga qo’shing.

$$2x^{2}-6+3x>0$$

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.

$$2x^{2}-6+3x=0$$

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni bu formula bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat tenglamada a uchun 2 ni, b uchun 3 ni va c uchun -6 ni ayiring.

$$x=\frac{-3±\sqrt{3^{2}-4\times 2\left(-6\right)}}{2\times 2}$$

Hisoblarni amalga oshiring.

$$x=\frac{-3±\sqrt{57}}{4}$$

x=\frac{-3±\sqrt{57}}{4} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$x=\frac{\sqrt{57}-3}{4} x=\frac{-\sqrt{57}-3}{4}$$

Yechimlardan foydalanib tengsizlikni qaytadan yozing.

$$2\left(x-\frac{\sqrt{57}-3}{4}\right)\left(x-\frac{-\sqrt{57}-3}{4}\right)>0$$

Koʻpaytma musbat boʻlishi uchun x-\frac{\sqrt{57}-3}{4} va x-\frac{-\sqrt{57}-3}{4} ikkalasi yo manfiy, yo musbat boʻlishi kerak. x-\frac{\sqrt{57}-3}{4} va x-\frac{-\sqrt{57}-3}{4} ikkalasi manfiy boʻlganda, yechimini toping.

$$x-\frac{\sqrt{57}-3}{4}<0 x-\frac{-\sqrt{57}-3}{4}<0$$

Ikkala tengsizlikning mos yechimi – x<\frac{-\sqrt{57}-3}{4}.

$$x<\frac{-\sqrt{57}-3}{4}$$

x-\frac{\sqrt{57}-3}{4} va x-\frac{-\sqrt{57}-3}{4} ikkalasi musbat boʻlganda, yechimini toping.

$$x-\frac{-\sqrt{57}-3}{4}>0 x-\frac{\sqrt{57}-3}{4}>0$$

Ikkala tengsizlikning mos yechimi – x>\frac{\sqrt{57}-3}{4}.

$$x>\frac{\sqrt{57}-3}{4}$$

Oxirgi yechim olingan yechimlarning birlashmasidir.

$$x<\frac{-\sqrt{57}-3}{4}\text{; }x>\frac{\sqrt{57}-3}{4}$$