s^2-3s=1 eching

$$s ^ { 2 } - 3 s = 1$$

s uchun yechish

$$s = \frac{\sqrt{13} + 3}{2} \approx 3,302775638$$

$$s=\frac{3-\sqrt{13}}{2}\approx -0,302775638$$

Kvadrat tenglamalaridan foydalanish qoidalari

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni kvadrat formulasi bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat formula ikki yechmni taqdim qiladi, biri ± qo'shish bo'lganda, va ikkinchisi ayiruv bo'lganda.

$$s^{2}-3s=1$$

Tenglamaning ikkala tarafidan 1 ni ayirish.

$$s^{2}-3s-1=1-1$$

O‘zidan 1 ayirilsa 0 qoladi.

$$s^{2}-3s-1=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, -3 ni b va -1 ni c bilan almashtiring.

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

-3 kvadratini chiqarish.

$$s=\frac{-\left(-3\right)±\sqrt{9-4\left(-1\right)}}{2}$$

-4 ni -1 marotabaga ko'paytirish.

$$s=\frac{-\left(-3\right)±\sqrt{9+4}}{2}$$

9 ni 4 ga qo'shish.

$$s=\frac{-\left(-3\right)±\sqrt{13}}{2}$$

-3 ning teskarisi 3 ga teng.

$$s=\frac{3±\sqrt{13}}{2}$$

s=\frac{3±\sqrt{13}}{2} tenglamasini yeching, bunda ± musbat. 3 ni \sqrt{13} ga qo'shish.

$$s=\frac{\sqrt{13}+3}{2}$$

s=\frac{3±\sqrt{13}}{2} tenglamasini yeching, bunda ± manfiy. 3 dan \sqrt{13} ni ayirish.

$$s=\frac{3-\sqrt{13}}{2}$$

Tenglama yechildi.

$$s=\frac{\sqrt{13}+3}{2} s=\frac{3-\sqrt{13}}{2}$$

Kvadratni yakunlash bo'yicha qadamlar

Bu kabi kvadrat tenglamalarni kvadratni yakunlab yechish mumkin. Kvadratni yechish uchun tenglama avval ushbu shaklda bo'lishi shart: x^{2}+bx=c.

$$s^{2}-3s=1$$

-3 ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{3}{2} olish uchun. Keyin, -\frac{3}{2} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$s^{2}-3s+\left(-\frac{3}{2}\right)^{2}=1+\left(-\frac{3}{2}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{3}{2} kvadratini chiqarish.

$$s^{2}-3s+\frac{9}{4}=1+\frac{9}{4}$$

1 ni \frac{9}{4} ga qo'shish.

$$s^{2}-3s+\frac{9}{4}=\frac{13}{4}$$

s^{2}-3s+\frac{9}{4} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(s-\frac{3}{2}\right)^{2}=\frac{13}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(s-\frac{3}{2}\right)^{2}}=\sqrt{\frac{13}{4}}$$

Qisqartirish.

$$s-\frac{3}{2}=\frac{\sqrt{13}}{2} s-\frac{3}{2}=-\frac{\sqrt{13}}{2}$$

\frac{3}{2} ni tenglamaning ikkala tarafiga qo'shish.

$$s=\frac{\sqrt{13}+3}{2} s=\frac{3-\sqrt{13}}{2}$$