6s^2-9s+1=0 eching

$$6 s ^ { 2 } - 9 s + 1 = 0$$

s uchun yechish

$$s=\frac{\sqrt{57}}{12}+\frac{3}{4}\approx 1,37915287$$

$$s=-\frac{\sqrt{57}}{12}+\frac{3}{4}\approx 0,12084713$$

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.

$$6s^{2}-9s+1=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 6 ni a, -9 ni b va 1 ni c bilan almashtiring.

$$s=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 6}}{2\times 6}$$

-9 kvadratini chiqarish.

$$s=\frac{-\left(-9\right)±\sqrt{81-4\times 6}}{2\times 6}$$

-4 ni 6 marotabaga ko'paytirish.

$$s=\frac{-\left(-9\right)±\sqrt{81-24}}{2\times 6}$$

81 ni -24 ga qo'shish.

$$s=\frac{-\left(-9\right)±\sqrt{57}}{2\times 6}$$

-9 ning teskarisi 9 ga teng.

$$s=\frac{9±\sqrt{57}}{2\times 6}$$

2 ni 6 marotabaga ko'paytirish.

$$s=\frac{9±\sqrt{57}}{12}$$

s=\frac{9±\sqrt{57}}{12} tenglamasini yeching, bunda ± musbat. 9 ni \sqrt{57} ga qo'shish.

$$s=\frac{\sqrt{57}+9}{12}$$

9+\sqrt{57} ni 12 ga bo'lish.

$$s=\frac{\sqrt{57}}{12}+\frac{3}{4}$$

s=\frac{9±\sqrt{57}}{12} tenglamasini yeching, bunda ± manfiy. 9 dan \sqrt{57} ni ayirish.

$$s=\frac{9-\sqrt{57}}{12}$$

9-\sqrt{57} ni 12 ga bo'lish.

$$s=-\frac{\sqrt{57}}{12}+\frac{3}{4}$$

Tenglama yechildi.

$$s=\frac{\sqrt{57}}{12}+\frac{3}{4} s=-\frac{\sqrt{57}}{12}+\frac{3}{4}$$

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.

$$6s^{2}-9s+1=0$$

Tenglamaning ikkala tarafidan 1 ni ayirish.

$$6s^{2}-9s+1-1=-1$$

O‘zidan 1 ayirilsa 0 qoladi.

$$6s^{2}-9s=-1$$

Ikki tarafini 6 ga bo‘ling.

$$\frac{6s^{2}-9s}{6}=-\frac{1}{6}$$

6 ga bo'lish 6 ga ko'paytirishni bekor qiladi.

$$s^{2}+\left(-\frac{9}{6}\right)s=-\frac{1}{6}$$

\frac{-9}{6} ulushini 3 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$s^{2}-\frac{3}{2}s=-\frac{1}{6}$$

-\frac{3}{2} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{3}{4} olish uchun. Keyin, -\frac{3}{4} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$s^{2}-\frac{3}{2}s+\left(-\frac{3}{4}\right)^{2}=-\frac{1}{6}+\left(-\frac{3}{4}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{3}{4} kvadratini chiqarish.

$$s^{2}-\frac{3}{2}s+\frac{9}{16}=-\frac{1}{6}+\frac{9}{16}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{1}{6} ni \frac{9}{16} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$s^{2}-\frac{3}{2}s+\frac{9}{16}=\frac{19}{48}$$

s^{2}-\frac{3}{2}s+\frac{9}{16} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(s-\frac{3}{4}\right)^{2}=\frac{19}{48}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(s-\frac{3}{4}\right)^{2}}=\sqrt{\frac{19}{48}}$$

Qisqartirish.

$$s-\frac{3}{4}=\frac{\sqrt{57}}{12} s-\frac{3}{4}=-\frac{\sqrt{57}}{12}$$

\frac{3}{4} ni tenglamaning ikkala tarafiga qo'shish.

$$s=\frac{\sqrt{57}}{12}+\frac{3}{4} s=-\frac{\sqrt{57}}{12}+\frac{3}{4}$$