6x^2-13x-5geq0 eching

$$6 x ^ { 2 } - 13 x - 5 \geq 0$$

x uchun yechish

$$x\in (-\infty,-\frac{1}{3}]\cup [\frac{5}{2},\infty)$$

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.

$$6x^{2}-13x-5=0$$

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni bu formula bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat tenglamada a uchun 6 ni, b uchun -13 ni va c uchun -5 ni ayiring.

$$x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 6\left(-5\right)}}{2\times 6}$$

Hisoblarni amalga oshiring.

$$x=\frac{13±17}{12}$$

x=\frac{13±17}{12} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$x=\frac{5}{2} x=-\frac{1}{3}$$

Yechimlardan foydalanib tengsizlikni qaytadan yozing.

$$6\left(x-\frac{5}{2}\right)\left(x+\frac{1}{3}\right)\geq 0$$

Koʻpaytma ≥0 boʻlishi uchun x-\frac{5}{2} va x+\frac{1}{3} ikkalasi ≤0 yoki ≥0 boʻlishi kerak. x-\frac{5}{2} va x+\frac{1}{3} ikkalasi ≤0 ga teng boʻlganda, yechimini toping.

$$x-\frac{5}{2}\leq 0 x+\frac{1}{3}\leq 0$$

Ikkala tengsizlikning mos yechimi – x\leq -\frac{1}{3}.

$$x\leq -\frac{1}{3}$$

x-\frac{5}{2} va x+\frac{1}{3} ikkalasi ≥0 ga teng boʻlganda, yechimini toping.

$$x+\frac{1}{3}\geq 0 x-\frac{5}{2}\geq 0$$

Ikkala tengsizlikning mos yechimi – x\geq \frac{5}{2}.

$$x\geq \frac{5}{2}$$

Oxirgi yechim olingan yechimlarning birlashmasidir.

$$x\leq -\frac{1}{3}\text{; }x\geq \frac{5}{2}$$