8x^2+8x+(-1)leq0 eching

$$8 x ^ { 2 } + 8 x + ( - 1 ) \leq 0$$

x uchun yechish

$$x\in \begin{bmatrix}-\frac{\sqrt{6}}{4}-\frac{1}{2},\frac{\sqrt{6}}{4}-\frac{1}{2}\end{bmatrix}$$

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.

$$8x^{2}+8x-1=0$$

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni bu formula bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat tenglamada a uchun 8 ni, b uchun 8 ni va c uchun -1 ni ayiring.

$$x=\frac{-8±\sqrt{8^{2}-4\times 8\left(-1\right)}}{2\times 8}$$

Hisoblarni amalga oshiring.

$$x=\frac{-8±4\sqrt{6}}{16}$$

x=\frac{-8±4\sqrt{6}}{16} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$x=\frac{\sqrt{6}}{4}-\frac{1}{2} x=-\frac{\sqrt{6}}{4}-\frac{1}{2}$$

Yechimlardan foydalanib tengsizlikni qaytadan yozing.

$$8\left(x-\left(\frac{\sqrt{6}}{4}-\frac{1}{2}\right)\right)\left(x-\left(-\frac{\sqrt{6}}{4}-\frac{1}{2}\right)\right)\leq 0$$

Koʻpaytma ≤0 boʻlishi uchun qiymatlardan biri x-\left(\frac{\sqrt{6}}{4}-\frac{1}{2}\right) va x-\left(-\frac{\sqrt{6}}{4}-\frac{1}{2}\right) ≥0 va boshqasi ≤0 boʻlishi kerak. x-\left(\frac{\sqrt{6}}{4}-\frac{1}{2}\right)\geq 0 va x-\left(-\frac{\sqrt{6}}{4}-\frac{1}{2}\right)\leq 0 boʻlgandagi holatni koʻrib chiqing.

$$x-\left(\frac{\sqrt{6}}{4}-\frac{1}{2}\right)\geq 0 x-\left(-\frac{\sqrt{6}}{4}-\frac{1}{2}\right)\leq 0$$

Bu har qanday x uchun xato.

$$x\in \emptyset$$

x-\left(\frac{\sqrt{6}}{4}-\frac{1}{2}\right)\leq 0 va x-\left(-\frac{\sqrt{6}}{4}-\frac{1}{2}\right)\geq 0 boʻlgandagi holatni koʻrib chiqing.

$$x-\left(-\frac{\sqrt{6}}{4}-\frac{1}{2}\right)\geq 0 x-\left(\frac{\sqrt{6}}{4}-\frac{1}{2}\right)\leq 0$$

Ikkala tengsizlikning mos yechimi – x\in \left[-\frac{\sqrt{6}}{4}-\frac{1}{2},\frac{\sqrt{6}}{4}-\frac{1}{2}\right].

$$x\in \begin{bmatrix}-\frac{\sqrt{6}}{4}-\frac{1}{2},\frac{\sqrt{6}}{4}-\frac{1}{2}\end{bmatrix}$$

Oxirgi yechim olingan yechimlarning birlashmasidir.

$$x\in \begin{bmatrix}-\frac{\sqrt{6}}{4}-\frac{1}{2},\frac{\sqrt{6}}{4}-\frac{1}{2}\end{bmatrix}$$