3x^2=x+10 eching

$$3 x ^ { 2 } = x + 10$$

x uchun yechish

$$x = -\frac{5}{3} = -1\frac{2}{3} \approx -1,666666667$$

$$x=2$$

Guruhlash orqali faktorlash qadamlari

Ikkala tarafdan x ni ayirish.

$$3x^{2}-x=10$$

Ikkala tarafdan 10 ni ayirish.

$$3x^{2}-x-10=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon 3x^{2}+ax+bx-10 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-1 ab=3\left(-10\right)=-30$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -30-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-30 2,-15 3,-10 5,-6$$

Har bir juftlik yigʻindisini hisoblang.

$$1-30=-29 2-15=-13 3-10=-7 5-6=-1$$

Yechim – -1 yigʻindisini beruvchi juftlik.

$$a=-6 b=5$$

3x^{2}-x-10 ni \left(3x^{2}-6x\right)+\left(5x-10\right) sifatida qaytadan yozish.

$$\left(3x^{2}-6x\right)+\left(5x-10\right)$$

Birinchi guruhda 3x ni va ikkinchi guruhda 5 ni faktordan chiqaring.

$$3x\left(x-2\right)+5\left(x-2\right)$$

Distributiv funktsiyasidan foydalangan holda x-2 umumiy terminini chiqaring.

$$\left(x-2\right)\left(3x+5\right)$$

Tenglamani yechish uchun x-2=0 va 3x+5=0 ni yeching.

$$x=2 x=-\frac{5}{3}$$

Kvadrat tenglamalaridan foydalanish qoidalari

Ikkala tarafdan x ni ayirish.

$$3x^{2}-x=10$$

Ikkala tarafdan 10 ni ayirish.

$$3x^{2}-x-10=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 3 ni a, -1 ni b va -10 ni c bilan almashtiring.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 3\left(-10\right)}}{2\times 3}$$

-4 ni 3 marotabaga ko'paytirish.

$$x=\frac{-\left(-1\right)±\sqrt{1-12\left(-10\right)}}{2\times 3}$$

-12 ni -10 marotabaga ko'paytirish.

$$x=\frac{-\left(-1\right)±\sqrt{1+120}}{2\times 3}$$

1 ni 120 ga qo'shish.

$$x=\frac{-\left(-1\right)±\sqrt{121}}{2\times 3}$$

121 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-1\right)±11}{2\times 3}$$

-1 ning teskarisi 1 ga teng.

$$x=\frac{1±11}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$x=\frac{1±11}{6}$$

x=\frac{1±11}{6} tenglamasini yeching, bunda ± musbat. 1 ni 11 ga qo'shish.

$$x=\frac{12}{6}$$

12 ni 6 ga bo'lish.

$$x=2$$

x=\frac{1±11}{6} tenglamasini yeching, bunda ± manfiy. 1 dan 11 ni ayirish.

$$x=-\frac{10}{6}$$

\frac{-10}{6} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=-\frac{5}{3}$$

Tenglama yechildi.

$$x=2 x=-\frac{5}{3}$$

Kvadratni yakunlash bo'yicha qadamlar

Ikkala tarafdan x ni ayirish.

$$3x^{2}-x=10$$

Ikki tarafini 3 ga bo‘ling.

$$\frac{3x^{2}-x}{3}=\frac{10}{3}$$

3 ga bo'lish 3 ga ko'paytirishni bekor qiladi.

$$x^{2}-\frac{1}{3}x=\frac{10}{3}$$

-\frac{1}{3} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{1}{6} olish uchun. Keyin, -\frac{1}{6} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-\frac{1}{3}x+\left(-\frac{1}{6}\right)^{2}=\frac{10}{3}+\left(-\frac{1}{6}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{1}{6} kvadratini chiqarish.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{10}{3}+\frac{1}{36}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{10}{3} ni \frac{1}{36} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{121}{36}$$

x^{2}-\frac{1}{3}x+\frac{1}{36} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-\frac{1}{6}\right)^{2}=\frac{121}{36}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{1}{6}\right)^{2}}=\sqrt{\frac{121}{36}}$$

Qisqartirish.

$$x-\frac{1}{6}=\frac{11}{6} x-\frac{1}{6}=-\frac{11}{6}$$

\frac{1}{6} ni tenglamaning ikkala tarafiga qo'shish.

$$x=2 x=-\frac{5}{3}$$