a^2-68a+225leq0 eching

$$a ^ { 2 } - 68 a + 225 \leq 0$$

a uchun yechish

$$a\in \begin{bmatrix}34-7\sqrt{19},7\sqrt{19}+34\end{bmatrix}$$

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.

$$a^{2}-68a+225=0$$

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 -68 ni va c uchun 225 ni ayiring.

$$a=\frac{-\left(-68\right)±\sqrt{\left(-68\right)^{2}-4\times 1\times 225}}{2}$$

Hisoblarni amalga oshiring.

$$a=\frac{68±14\sqrt{19}}{2}$$

a=\frac{68±14\sqrt{19}}{2} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$a=7\sqrt{19}+34 a=34-7\sqrt{19}$$

Yechimlardan foydalanib tengsizlikni qaytadan yozing.

$$\left(a-\left(7\sqrt{19}+34\right)\right)\left(a-\left(34-7\sqrt{19}\right)\right)\leq 0$$

Koʻpaytma ≤0 boʻlishi uchun qiymatlardan biri a-\left(7\sqrt{19}+34\right) va a-\left(34-7\sqrt{19}\right) ≥0 va boshqasi ≤0 boʻlishi kerak. Consider the case when a-\left(7\sqrt{19}+34\right)\geq 0 and a-\left(34-7\sqrt{19}\right)\leq 0.

$$a-\left(7\sqrt{19}+34\right)\geq 0 a-\left(34-7\sqrt{19}\right)\leq 0$$

Bu har qanday a uchun xato.

$$a\in \emptyset$$

Consider the case when a-\left(7\sqrt{19}+34\right)\leq 0 and a-\left(34-7\sqrt{19}\right)\geq 0.

$$a-\left(34-7\sqrt{19}\right)\geq 0 a-\left(7\sqrt{19}+34\right)\leq 0$$

Ikkala tengsizlikning mos yechimi – a\in \left[34-7\sqrt{19},7\sqrt{19}+34\right].

$$a\in \begin{bmatrix}34-7\sqrt{19},7\sqrt{19}+34\end{bmatrix}$$

Oxirgi yechim olingan yechimlarning birlashmasidir.

$$a\in \begin{bmatrix}34-7\sqrt{19},7\sqrt{19}+34\end{bmatrix}$$