x^2+11x-10geq0 eching

$${ x }^{ 2 } +11x-10 \geq 0$$

x uchun yechish

$$x\in (-\infty,\frac{-\sqrt{161}-11}{2}]\cup [\frac{\sqrt{161}-11}{2},\infty)$$

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.

$$x^{2}+11x-10=0$$

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 11 ni va c uchun -10 ni ayiring.

$$x=\frac{-11±\sqrt{11^{2}-4\times 1\left(-10\right)}}{2}$$

Hisoblarni amalga oshiring.

$$x=\frac{-11±\sqrt{161}}{2}$$

x=\frac{-11±\sqrt{161}}{2} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$x=\frac{\sqrt{161}-11}{2} x=\frac{-\sqrt{161}-11}{2}$$

Yechimlardan foydalanib tengsizlikni qaytadan yozing.

$$\left(x-\frac{\sqrt{161}-11}{2}\right)\left(x-\frac{-\sqrt{161}-11}{2}\right)\geq 0$$

Koʻpaytma ≥0 boʻlishi uchun x-\frac{\sqrt{161}-11}{2} va x-\frac{-\sqrt{161}-11}{2} ikkalasi ≤0 yoki ≥0 boʻlishi kerak. x-\frac{\sqrt{161}-11}{2} va x-\frac{-\sqrt{161}-11}{2} ikkalasi ≤0 ga teng boʻlganda, yechimini toping.

$$x-\frac{\sqrt{161}-11}{2}\leq 0 x-\frac{-\sqrt{161}-11}{2}\leq 0$$

Ikkala tengsizlikning mos yechimi – x\leq \frac{-\sqrt{161}-11}{2}.

$$x\leq \frac{-\sqrt{161}-11}{2}$$

x-\frac{\sqrt{161}-11}{2} va x-\frac{-\sqrt{161}-11}{2} ikkalasi ≥0 ga teng boʻlganda, yechimini toping.

$$x-\frac{-\sqrt{161}-11}{2}\geq 0 x-\frac{\sqrt{161}-11}{2}\geq 0$$

Ikkala tengsizlikning mos yechimi – x\geq \frac{\sqrt{161}-11}{2}.

$$x\geq \frac{\sqrt{161}-11}{2}$$

Oxirgi yechim olingan yechimlarning birlashmasidir.

$$x\leq \frac{-\sqrt{161}-11}{2}\text{; }x\geq \frac{\sqrt{161}-11}{2}$$