x^2+36x+240>0 eching

$$x ^ { 2 } + 36 x + 240 > 0$$

x uchun yechish

$$x\in \left(-\infty,-2\sqrt{21}-18\right)\cup \left(2\sqrt{21}-18,\infty\right)$$

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.

$$x^{2}+36x+240=0$$

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 36 ni va c uchun 240 ni ayiring.

$$x=\frac{-36±\sqrt{36^{2}-4\times 1\times 240}}{2}$$

Hisoblarni amalga oshiring.

$$x=\frac{-36±4\sqrt{21}}{2}$$

x=\frac{-36±4\sqrt{21}}{2} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$x=2\sqrt{21}-18 x=-2\sqrt{21}-18$$

Yechimlardan foydalanib tengsizlikni qaytadan yozing.

$$\left(x-\left(2\sqrt{21}-18\right)\right)\left(x-\left(-2\sqrt{21}-18\right)\right)>0$$

Koʻpaytma musbat boʻlishi uchun x-\left(2\sqrt{21}-18\right) va x-\left(-2\sqrt{21}-18\right) ikkalasi yo manfiy, yo musbat boʻlishi kerak. x-\left(2\sqrt{21}-18\right) va x-\left(-2\sqrt{21}-18\right) ikkalasi manfiy boʻlganda, yechimini toping.

$$x-\left(2\sqrt{21}-18\right)<0 x-\left(-2\sqrt{21}-18\right)<0$$

Ikkala tengsizlikning mos yechimi – x<-2\sqrt{21}-18.

$$x<-2\sqrt{21}-18$$

x-\left(2\sqrt{21}-18\right) va x-\left(-2\sqrt{21}-18\right) ikkalasi musbat boʻlganda, yechimini toping.

$$x-\left(-2\sqrt{21}-18\right)>0 x-\left(2\sqrt{21}-18\right)>0$$

Ikkala tengsizlikning mos yechimi – x>2\sqrt{21}-18.

$$x>2\sqrt{21}-18$$

Oxirgi yechim olingan yechimlarning birlashmasidir.

$$x<-2\sqrt{21}-18\text{; }x>2\sqrt{21}-18$$