3x^2=22x-7 eching

$$3 x ^ { 2 } = 22 x - 7$$

x uchun yechish

$$x=\frac{1}{3}\approx 0,333333333$$

$$x=7$$

Guruhlash orqali faktorlash qadamlari

Ikkala tarafdan 22x ni ayirish.

$$3x^{2}-22x=-7$$

7 ni ikki tarafga qo’shing.

$$3x^{2}-22x+7=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon 3x^{2}+ax+bx+7 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-22 ab=3\times 7=21$$

ab musbat boʻlganda, a va b da bir xil belgi bor. a+b manfiy boʻlganda, a va b ikkisi ham manfiy. 21-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,-21 -3,-7$$

Har bir juftlik yigʻindisini hisoblang.

$$-1-21=-22 -3-7=-10$$

Yechim – -22 yigʻindisini beruvchi juftlik.

$$a=-21 b=-1$$

3x^{2}-22x+7 ni \left(3x^{2}-21x\right)+\left(-x+7\right) sifatida qaytadan yozish.

$$\left(3x^{2}-21x\right)+\left(-x+7\right)$$

Birinchi guruhda 3x ni va ikkinchi guruhda -1 ni faktordan chiqaring.

$$3x\left(x-7\right)-\left(x-7\right)$$

Distributiv funktsiyasidan foydalangan holda x-7 umumiy terminini chiqaring.

$$\left(x-7\right)\left(3x-1\right)$$

Tenglamani yechish uchun x-7=0 va 3x-1=0 ni yeching.

$$x=7 x=\frac{1}{3}$$

Kvadrat tenglamalaridan foydalanish qoidalari

Ikkala tarafdan 22x ni ayirish.

$$3x^{2}-22x=-7$$

7 ni ikki tarafga qo’shing.

$$3x^{2}-22x+7=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 3 ni a, -22 ni b va 7 ni c bilan almashtiring.

$$x=\frac{-\left(-22\right)±\sqrt{\left(-22\right)^{2}-4\times 3\times 7}}{2\times 3}$$

-22 kvadratini chiqarish.

$$x=\frac{-\left(-22\right)±\sqrt{484-4\times 3\times 7}}{2\times 3}$$

-4 ni 3 marotabaga ko'paytirish.

$$x=\frac{-\left(-22\right)±\sqrt{484-12\times 7}}{2\times 3}$$

-12 ni 7 marotabaga ko'paytirish.

$$x=\frac{-\left(-22\right)±\sqrt{484-84}}{2\times 3}$$

484 ni -84 ga qo'shish.

$$x=\frac{-\left(-22\right)±\sqrt{400}}{2\times 3}$$

400 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-22\right)±20}{2\times 3}$$

-22 ning teskarisi 22 ga teng.

$$x=\frac{22±20}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$x=\frac{22±20}{6}$$

x=\frac{22±20}{6} tenglamasini yeching, bunda ± musbat. 22 ni 20 ga qo'shish.

$$x=\frac{42}{6}$$

42 ni 6 ga bo'lish.

$$x=7$$

x=\frac{22±20}{6} tenglamasini yeching, bunda ± manfiy. 22 dan 20 ni ayirish.

$$x=\frac{2}{6}$$

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

$$x=\frac{1}{3}$$

Tenglama yechildi.

$$x=7 x=\frac{1}{3}$$

Kvadratni yakunlash bo'yicha qadamlar

Ikkala tarafdan 22x ni ayirish.

$$3x^{2}-22x=-7$$

Ikki tarafini 3 ga bo‘ling.

$$\frac{3x^{2}-22x}{3}=-\frac{7}{3}$$

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

$$x^{2}-\frac{22}{3}x=-\frac{7}{3}$$

-\frac{22}{3} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{11}{3} olish uchun. Keyin, -\frac{11}{3} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-\frac{22}{3}x+\left(-\frac{11}{3}\right)^{2}=-\frac{7}{3}+\left(-\frac{11}{3}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{11}{3} kvadratini chiqarish.

$$x^{2}-\frac{22}{3}x+\frac{121}{9}=-\frac{7}{3}+\frac{121}{9}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{7}{3} ni \frac{121}{9} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}-\frac{22}{3}x+\frac{121}{9}=\frac{100}{9}$$

x^{2}-\frac{22}{3}x+\frac{121}{9} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-\frac{11}{3}\right)^{2}=\frac{100}{9}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{11}{3}\right)^{2}}=\sqrt{\frac{100}{9}}$$

Qisqartirish.

$$x-\frac{11}{3}=\frac{10}{3} x-\frac{11}{3}=-\frac{10}{3}$$

\frac{11}{3} ni tenglamaning ikkala tarafiga qo'shish.

$$x=7 x=\frac{1}{3}$$