x^2+14x-28leq0 eching

$$x ^ { 2 } + 14 x - 28 \leq 0$$

x uchun yechish

$$x\in \begin{bmatrix}-\sqrt{77}-7,\sqrt{77}-7\end{bmatrix}$$

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.

$$x^{2}+14x-28=0$$

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 14 ni va c uchun -28 ni ayiring.

$$x=\frac{-14±\sqrt{14^{2}-4\times 1\left(-28\right)}}{2}$$

Hisoblarni amalga oshiring.

$$x=\frac{-14±2\sqrt{77}}{2}$$

x=\frac{-14±2\sqrt{77}}{2} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$x=\sqrt{77}-7 x=-\sqrt{77}-7$$

Yechimlardan foydalanib tengsizlikni qaytadan yozing.

$$\left(x-\left(\sqrt{77}-7\right)\right)\left(x-\left(-\sqrt{77}-7\right)\right)\leq 0$$

Koʻpaytma ≤0 boʻlishi uchun qiymatlardan biri x-\left(\sqrt{77}-7\right) va x-\left(-\sqrt{77}-7\right) ≥0 va boshqasi ≤0 boʻlishi kerak. Consider the case when x-\left(\sqrt{77}-7\right)\geq 0 and x-\left(-\sqrt{77}-7\right)\leq 0.

$$x-\left(\sqrt{77}-7\right)\geq 0 x-\left(-\sqrt{77}-7\right)\leq 0$$

Bu har qanday x uchun xato.

$$x\in \emptyset$$

Consider the case when x-\left(\sqrt{77}-7\right)\leq 0 and x-\left(-\sqrt{77}-7\right)\geq 0.

$$x-\left(-\sqrt{77}-7\right)\geq 0 x-\left(\sqrt{77}-7\right)\leq 0$$

Ikkala tengsizlikning mos yechimi – x\in \left[-\left(\sqrt{77}+7\right),\sqrt{77}-7\right].

$$x\in \begin{bmatrix}-\left(\sqrt{77}+7\right),\sqrt{77}-7\end{bmatrix}$$

Oxirgi yechim olingan yechimlarning birlashmasidir.

$$x\in \begin{bmatrix}-\sqrt{77}-7,\sqrt{77}-7\end{bmatrix}$$