x^2+4x-7<0 eching

$$x ^ { 2 } + 4 x - 7 < 0$$

x uchun yechish

$$x\in \left(-\sqrt{11}-2,\sqrt{11}-2\right)$$

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.

$$x^{2}+4x-7=0$$

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 4 ni va c uchun -7 ni ayiring.

$$x=\frac{-4±\sqrt{4^{2}-4\times 1\left(-7\right)}}{2}$$

Hisoblarni amalga oshiring.

$$x=\frac{-4±2\sqrt{11}}{2}$$

x=\frac{-4±2\sqrt{11}}{2} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$x=\sqrt{11}-2 x=-\sqrt{11}-2$$

Yechimlardan foydalanib tengsizlikni qaytadan yozing.

$$\left(x-\left(\sqrt{11}-2\right)\right)\left(x-\left(-\sqrt{11}-2\right)\right)<0$$

Koʻpaytma manfiy boʻlishi uchun x-\left(\sqrt{11}-2\right) va x-\left(-\sqrt{11}-2\right) qarama-qarshi belgilar boʻlishi kerak. x-\left(\sqrt{11}-2\right) musbat, x-\left(-\sqrt{11}-2\right) manfiy boʻlganda, yechimni toping.

$$x-\left(\sqrt{11}-2\right)>0 x-\left(-\sqrt{11}-2\right)<0$$

Bu har qanday x uchun xato.

$$x\in \emptyset$$

x-\left(-\sqrt{11}-2\right) musbat, x-\left(\sqrt{11}-2\right) manfiy boʻlganda, yechimni toping.

$$x-\left(-\sqrt{11}-2\right)>0 x-\left(\sqrt{11}-2\right)<0$$

Ikkala tengsizlikning mos yechimi – x\in \left(-\left(\sqrt{11}+2\right),\sqrt{11}-2\right).

$$x\in \left(-\left(\sqrt{11}+2\right),\sqrt{11}-2\right)$$

Oxirgi yechim olingan yechimlarning birlashmasidir.

$$x\in \left(-\sqrt{11}-2,\sqrt{11}-2\right)$$