a^2+3a-60>0 eching

$$a ^ { 2 } + 3 a - 60 > 0$$

a uchun yechish

$$a\in \left(-\infty,\frac{-\sqrt{249}-3}{2}\right)\cup \left(\frac{\sqrt{249}-3}{2},\infty\right)$$

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.

$$a^{2}+3a-60=0$$

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 3 ni va c uchun -60 ni ayiring.

$$a=\frac{-3±\sqrt{3^{2}-4\times 1\left(-60\right)}}{2}$$

Hisoblarni amalga oshiring.

$$a=\frac{-3±\sqrt{249}}{2}$$

a=\frac{-3±\sqrt{249}}{2} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$a=\frac{\sqrt{249}-3}{2} a=\frac{-\sqrt{249}-3}{2}$$

Yechimlardan foydalanib tengsizlikni qaytadan yozing.

$$\left(a-\frac{\sqrt{249}-3}{2}\right)\left(a-\frac{-\sqrt{249}-3}{2}\right)>0$$

Koʻpaytma musbat boʻlishi uchun a-\frac{\sqrt{249}-3}{2} va a-\frac{-\sqrt{249}-3}{2} ikkalasi yo manfiy, yo musbat boʻlishi kerak. a-\frac{\sqrt{249}-3}{2} va a-\frac{-\sqrt{249}-3}{2} ikkalasi manfiy boʻlganda, yechimini toping.

$$a-\frac{\sqrt{249}-3}{2}<0 a-\frac{-\sqrt{249}-3}{2}<0$$

Ikkala tengsizlikning mos yechimi – a<\frac{-\sqrt{249}-3}{2}.

$$a<\frac{-\sqrt{249}-3}{2}$$

a-\frac{\sqrt{249}-3}{2} va a-\frac{-\sqrt{249}-3}{2} ikkalasi musbat boʻlganda, yechimini toping.

$$a-\frac{-\sqrt{249}-3}{2}>0 a-\frac{\sqrt{249}-3}{2}>0$$

Ikkala tengsizlikning mos yechimi – a>\frac{\sqrt{249}-3}{2}.

$$a>\frac{\sqrt{249}-3}{2}$$

Oxirgi yechim olingan yechimlarning birlashmasidir.

$$a<\frac{-\sqrt{249}-3}{2}\text{; }a>\frac{\sqrt{249}-3}{2}$$