m^2-m-3/4geq0 eching

$$m ^ { 2 } - m - \frac { 3 } { 4 } \geq 0$$

m uchun yechish

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

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.

$$m^{2}-m-\frac{3}{4}=0$$

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 -1 ni va c uchun -\frac{3}{4} ni ayiring.

$$m=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 1\left(-\frac{3}{4}\right)}}{2}$$

Hisoblarni amalga oshiring.

$$m=\frac{1±2}{2}$$

m=\frac{1±2}{2} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$m=\frac{3}{2} m=-\frac{1}{2}$$

Yechimlardan foydalanib tengsizlikni qaytadan yozing.

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

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

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

Ikkala tengsizlikning mos yechimi – m\leq -\frac{1}{2}.

$$m\leq -\frac{1}{2}$$

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

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

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

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

Oxirgi yechim olingan yechimlarning birlashmasidir.

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