6x^2-5x-6=0 eching

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

x uchun yechish

$$x=-\frac{2}{3}\approx -0,666666667$$

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

Guruhlash orqali faktorlash qadamlari

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon 6x^{2}+ax+bx-6 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-5 ab=6\left(-6\right)=-36$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -36-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-36 2,-18 3,-12 4,-9 6,-6$$

Har bir juftlik yigʻindisini hisoblang.

$$1-36=-35 2-18=-16 3-12=-9 4-9=-5 6-6=0$$

Yechim – -5 yigʻindisini beruvchi juftlik.

$$a=-9 b=4$$

6x^{2}-5x-6 ni \left(6x^{2}-9x\right)+\left(4x-6\right) sifatida qaytadan yozish.

$$\left(6x^{2}-9x\right)+\left(4x-6\right)$$

Birinchi guruhda 3x ni va ikkinchi guruhda 2 ni faktordan chiqaring.

$$3x\left(2x-3\right)+2\left(2x-3\right)$$

Distributiv funktsiyasidan foydalangan holda 2x-3 umumiy terminini chiqaring.

$$\left(2x-3\right)\left(3x+2\right)$$

Tenglamani yechish uchun 2x-3=0 va 3x+2=0 ni yeching.

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

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.

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

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 6 ni a, -5 ni b va -6 ni c bilan almashtiring.

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

-5 kvadratini chiqarish.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 6\left(-6\right)}}{2\times 6}$$

-4 ni 6 marotabaga ko'paytirish.

$$x=\frac{-\left(-5\right)±\sqrt{25-24\left(-6\right)}}{2\times 6}$$

-24 ni -6 marotabaga ko'paytirish.

$$x=\frac{-\left(-5\right)±\sqrt{25+144}}{2\times 6}$$

25 ni 144 ga qo'shish.

$$x=\frac{-\left(-5\right)±\sqrt{169}}{2\times 6}$$

169 ning kvadrat ildizini chiqarish.

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

-5 ning teskarisi 5 ga teng.

$$x=\frac{5±13}{2\times 6}$$

2 ni 6 marotabaga ko'paytirish.

$$x=\frac{5±13}{12}$$

x=\frac{5±13}{12} tenglamasini yeching, bunda ± musbat. 5 ni 13 ga qo'shish.

$$x=\frac{18}{12}$$

\frac{18}{12} ulushini 6 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=\frac{3}{2}$$

x=\frac{5±13}{12} tenglamasini yeching, bunda ± manfiy. 5 dan 13 ni ayirish.

$$x=-\frac{8}{12}$$

\frac{-8}{12} ulushini 4 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

Tenglama yechildi.

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

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.

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

6 ni tenglamaning ikkala tarafiga qo'shish.

$$6x^{2}-5x-6-\left(-6\right)=-\left(-6\right)$$

O‘zidan -6 ayirilsa 0 qoladi.

$$6x^{2}-5x=-\left(-6\right)$$

0 dan -6 ni ayirish.

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

Ikki tarafini 6 ga bo‘ling.

$$\frac{6x^{2}-5x}{6}=\frac{6}{6}$$

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

$$x^{2}-\frac{5}{6}x=\frac{6}{6}$$

6 ni 6 ga bo'lish.

$$x^{2}-\frac{5}{6}x=1$$

-\frac{5}{6} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{5}{12} olish uchun. Keyin, -\frac{5}{12} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-\frac{5}{6}x+\left(-\frac{5}{12}\right)^{2}=1+\left(-\frac{5}{12}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{5}{12} kvadratini chiqarish.

$$x^{2}-\frac{5}{6}x+\frac{25}{144}=1+\frac{25}{144}$$

1 ni \frac{25}{144} ga qo'shish.

$$x^{2}-\frac{5}{6}x+\frac{25}{144}=\frac{169}{144}$$

x^{2}-\frac{5}{6}x+\frac{25}{144} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-\frac{5}{12}\right)^{2}=\frac{169}{144}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{5}{12}\right)^{2}}=\sqrt{\frac{169}{144}}$$

Qisqartirish.

$$x-\frac{5}{12}=\frac{13}{12} x-\frac{5}{12}=-\frac{13}{12}$$

\frac{5}{12} ni tenglamaning ikkala tarafiga qo'shish.

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