772-2x^2+x-1leq0 eching

$$772 - 2 x ^ { 2 } + x - 1 \leq 0$$

x uchun yechish

$$x\in (-\infty,\frac{1-\sqrt{6169}}{4}]\cup [\frac{\sqrt{6169}+1}{4},\infty)$$

Kvadrat tenglamalaridan foydalanish qoidalari

771 olish uchun 772 dan 1 ni ayirish.

$$771-2x^{2}+x\leq 0$$

771-2x^{2}+x musbatida eng katta quvvatni koeffitsientini aniqlash uchun tengsizlikni -1 ga koʻpaytiring. -1 manfiy boʻlgani uchun tengsizlikning yo‘nalishi o‘zgaradi.

$$-771+2x^{2}-x\geq 0$$

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.

$$-771+2x^{2}-x=0$$

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni bu formula bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat tenglamada a uchun 2 ni, b uchun -1 ni va c uchun -771 ni ayiring.

$$x=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 2\left(-771\right)}}{2\times 2}$$

Hisoblarni amalga oshiring.

$$x=\frac{1±\sqrt{6169}}{4}$$

x=\frac{1±\sqrt{6169}}{4} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$x=\frac{\sqrt{6169}+1}{4} x=\frac{1-\sqrt{6169}}{4}$$

Yechimlardan foydalanib tengsizlikni qaytadan yozing.

$$2\left(x-\frac{\sqrt{6169}+1}{4}\right)\left(x-\frac{1-\sqrt{6169}}{4}\right)\geq 0$$

Koʻpaytma ≥0 boʻlishi uchun x-\frac{\sqrt{6169}+1}{4} va x-\frac{1-\sqrt{6169}}{4} ikkalasi ≤0 yoki ≥0 boʻlishi kerak. x-\frac{\sqrt{6169}+1}{4} va x-\frac{1-\sqrt{6169}}{4} ikkalasi ≤0 ga teng boʻlganda, yechimini toping.

$$x-\frac{\sqrt{6169}+1}{4}\leq 0 x-\frac{1-\sqrt{6169}}{4}\leq 0$$

Ikkala tengsizlikning mos yechimi – x\leq \frac{1-\sqrt{6169}}{4}.

$$x\leq \frac{1-\sqrt{6169}}{4}$$

x-\frac{\sqrt{6169}+1}{4} va x-\frac{1-\sqrt{6169}}{4} ikkalasi ≥0 ga teng boʻlganda, yechimini toping.

$$x-\frac{1-\sqrt{6169}}{4}\geq 0 x-\frac{\sqrt{6169}+1}{4}\geq 0$$

Ikkala tengsizlikning mos yechimi – x\geq \frac{\sqrt{6169}+1}{4}.

$$x\geq \frac{\sqrt{6169}+1}{4}$$

Oxirgi yechim olingan yechimlarning birlashmasidir.

$$x\leq \frac{1-\sqrt{6169}}{4}\text{; }x\geq \frac{\sqrt{6169}+1}{4}$$