-2x^2+12x-14>0 eching

$$- 2 x ^ { 2 } + 12 x - 14 > 0$$

x uchun yechish

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

Kvadrat tenglamalaridan foydalanish qoidalari

-2x^{2}+12x-14 musbatida eng katta quvvatni koeffitsientini aniqlash uchun tengsizlikni -1 ga koʻpaytiring. -1 <0 bo‘lganda, tengsizlikning yo‘nalishi o‘zgargan bo‘ladi.

$$2x^{2}-12x+14<0$$

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.

$$2x^{2}-12x+14=0$$

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 -12 ni va c uchun 14 ni ayiring.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 2\times 14}}{2\times 2}$$

Hisoblarni amalga oshiring.

$$x=\frac{12±4\sqrt{2}}{4}$$

x=\frac{12±4\sqrt{2}}{4} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

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

Yechimlardan foydalanib tengsizlikni qaytadan yozing.

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

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.

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