x^2-6x+2>0 eching

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

x uchun yechish

$$x\in \left(-\infty,3-\sqrt{7}\right)\cup \left(\sqrt{7}+3,\infty\right)$$

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.

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

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 -6 ni va c uchun 2 ni ayiring.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 1\times 2}}{2}$$

Hisoblarni amalga oshiring.

$$x=\frac{6±2\sqrt{7}}{2}$$

x=\frac{6±2\sqrt{7}}{2} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$x=\sqrt{7}+3 x=3-\sqrt{7}$$

Yechimlardan foydalanib tengsizlikni qaytadan yozing.

$$\left(x-\left(\sqrt{7}+3\right)\right)\left(x-\left(3-\sqrt{7}\right)\right)>0$$

Koʻpaytma musbat boʻlishi uchun x-\left(\sqrt{7}+3\right) va x-\left(3-\sqrt{7}\right) ikkalasi yo manfiy, yo musbat boʻlishi kerak. x-\left(\sqrt{7}+3\right) va x-\left(3-\sqrt{7}\right) ikkalasi manfiy boʻlganda, yechimini toping.

$$x-\left(\sqrt{7}+3\right)<0 x-\left(3-\sqrt{7}\right)<0$$

Ikkala tengsizlikning mos yechimi – x<3-\sqrt{7}.

$$x<3-\sqrt{7}$$

x-\left(\sqrt{7}+3\right) va x-\left(3-\sqrt{7}\right) ikkalasi musbat boʻlganda, yechimini toping.

$$x-\left(3-\sqrt{7}\right)>0 x-\left(\sqrt{7}+3\right)>0$$

Ikkala tengsizlikning mos yechimi – x>\sqrt{7}+3.

$$x>\sqrt{7}+3$$

Oxirgi yechim olingan yechimlarning birlashmasidir.

$$x<3-\sqrt{7}\text{; }x>\sqrt{7}+3$$