10x^2-7x-12geq0 eching

$$10 x ^ { 2 } - 7 x - 12 \geq 0$$

x uchun yechish

$$x\in (-\infty,-\frac{4}{5}]\cup [\frac{3}{2},\infty)$$

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.

$$10x^{2}-7x-12=0$$

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni bu formula bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat tenglamada a uchun 10 ni, b uchun -7 ni va c uchun -12 ni ayiring.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 10\left(-12\right)}}{2\times 10}$$

Hisoblarni amalga oshiring.

$$x=\frac{7±23}{20}$$

x=\frac{7±23}{20} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$x=\frac{3}{2} x=-\frac{4}{5}$$

Yechimlardan foydalanib tengsizlikni qaytadan yozing.

$$10\left(x-\frac{3}{2}\right)\left(x+\frac{4}{5}\right)\geq 0$$

Koʻpaytma ≥0 boʻlishi uchun x-\frac{3}{2} va x+\frac{4}{5} ikkalasi ≤0 yoki ≥0 boʻlishi kerak. x-\frac{3}{2} va x+\frac{4}{5} ikkalasi ≤0 ga teng boʻlganda, yechimini toping.

$$x-\frac{3}{2}\leq 0 x+\frac{4}{5}\leq 0$$

Ikkala tengsizlikning mos yechimi – x\leq -\frac{4}{5}.

$$x\leq -\frac{4}{5}$$

x-\frac{3}{2} va x+\frac{4}{5} ikkalasi ≥0 ga teng boʻlganda, yechimini toping.

$$x+\frac{4}{5}\geq 0 x-\frac{3}{2}\geq 0$$

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

$$x\geq \frac{3}{2}$$

Oxirgi yechim olingan yechimlarning birlashmasidir.

$$x\leq -\frac{4}{5}\text{; }x\geq \frac{3}{2}$$