(3x-5)(x+2)leqx^2 eching

$$( 3 x - 5 ) ( x + 2 ) \leq x ^ { 2 }$$

x uchun yechish

$$x\in \begin{bmatrix}-\frac{5}{2},2\end{bmatrix}$$

Kvadrat tenglamalaridan foydalanish qoidalari

3x-5 ga x+2 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$3x^{2}+x-10\leq x^{2}$$

Ikkala tarafdan x^{2} ni ayirish.

$$3x^{2}+x-10-x^{2}\leq 0$$

2x^{2} ni olish uchun 3x^{2} va -x^{2} ni birlashtirish.

$$2x^{2}+x-10\leq 0$$

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.

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

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni bu formula bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat tenglamada a uchun 2 ni, b uchun 1 ni va c uchun -10 ni ayiring.

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

Hisoblarni amalga oshiring.

$$x=\frac{-1±9}{4}$$

x=\frac{-1±9}{4} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$x=2 x=-\frac{5}{2}$$

Yechimlardan foydalanib tengsizlikni qaytadan yozing.

$$2\left(x-2\right)\left(x+\frac{5}{2}\right)\leq 0$$

Koʻpaytma ≤0 boʻlishi uchun qiymatlardan biri x-2 va x+\frac{5}{2} ≥0 va boshqasi ≤0 boʻlishi kerak. Consider the case when x-2\geq 0 and x+\frac{5}{2}\leq 0.

$$x-2\geq 0 x+\frac{5}{2}\leq 0$$

Bu har qanday x uchun xato.

$$x\in \emptyset$$

Consider the case when x-2\leq 0 and x+\frac{5}{2}\geq 0.

$$x+\frac{5}{2}\geq 0 x-2\leq 0$$

Ikkala tengsizlikning mos yechimi – x\in \left[-\frac{5}{2},2\right].

$$x\in \begin{bmatrix}-\frac{5}{2},2\end{bmatrix}$$

Oxirgi yechim olingan yechimlarning birlashmasidir.

$$x\in \begin{bmatrix}-\frac{5}{2},2\end{bmatrix}$$