(2x)^2-12(x+1)geq0 eching

$$( 2 x ) ^ { 2 } - 12 ( x + 1 ) \geq 0$$

x uchun yechish

$$x\in (-\infty,\frac{3-\sqrt{21}}{2}]\cup [\frac{\sqrt{21}+3}{2},\infty)$$

Kvadrat tenglamalaridan foydalanish qoidalari

\left(2x\right)^{2} ni kengaytirish.

$$2^{2}x^{2}-12\left(x+1\right)\geq 0$$

2 daraja ko‘rsatkichini 2 ga hisoblang va 4 ni qiymatni oling.

$$4x^{2}-12\left(x+1\right)\geq 0$$

-12 ga x+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$4x^{2}-12x-12\geq 0$$

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.

$$4x^{2}-12x-12=0$$

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni bu formula bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat tenglamada a uchun 4 ni, b uchun -12 ni va c uchun -12 ni ayiring.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 4\left(-12\right)}}{2\times 4}$$

Hisoblarni amalga oshiring.

$$x=\frac{12±4\sqrt{21}}{8}$$

x=\frac{12±4\sqrt{21}}{8} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$x=\frac{\sqrt{21}+3}{2} x=\frac{3-\sqrt{21}}{2}$$

Yechimlardan foydalanib tengsizlikni qaytadan yozing.

$$4\left(x-\frac{\sqrt{21}+3}{2}\right)\left(x-\frac{3-\sqrt{21}}{2}\right)\geq 0$$

Koʻpaytma ≥0 boʻlishi uchun x-\frac{\sqrt{21}+3}{2} va x-\frac{3-\sqrt{21}}{2} ikkalasi ≤0 yoki ≥0 boʻlishi kerak. x-\frac{\sqrt{21}+3}{2} va x-\frac{3-\sqrt{21}}{2} ikkalasi ≤0 ga teng boʻlganda, yechimini toping.

$$x-\frac{\sqrt{21}+3}{2}\leq 0 x-\frac{3-\sqrt{21}}{2}\leq 0$$

Ikkala tengsizlikning mos yechimi – x\leq \frac{3-\sqrt{21}}{2}.

$$x\leq \frac{3-\sqrt{21}}{2}$$

x-\frac{\sqrt{21}+3}{2} va x-\frac{3-\sqrt{21}}{2} ikkalasi ≥0 ga teng boʻlganda, yechimini toping.

$$x-\frac{3-\sqrt{21}}{2}\geq 0 x-\frac{\sqrt{21}+3}{2}\geq 0$$

Ikkala tengsizlikning mos yechimi – x\geq \frac{\sqrt{21}+3}{2}.

$$x\geq \frac{\sqrt{21}+3}{2}$$

Oxirgi yechim olingan yechimlarning birlashmasidir.

$$x\leq \frac{3-\sqrt{21}}{2}\text{; }x\geq \frac{\sqrt{21}+3}{2}$$