-3x^2-17x=22 eching

$$- 3 x ^ { 2 } - 17 x = 22$$

x uchun yechish

$$x = -\frac{11}{3} = -3\frac{2}{3} \approx -3,666666667$$

$$x=-2$$

Guruhlash orqali faktorlash qadamlari

Ikkala tarafdan 22 ni ayirish.

$$-3x^{2}-17x-22=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon -3x^{2}+ax+bx-22 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-17 ab=-3\left(-22\right)=66$$

ab musbat boʻlganda, a va b da bir xil belgi bor. a+b manfiy boʻlganda, a va b ikkisi ham manfiy. 66-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,-66 -2,-33 -3,-22 -6,-11$$

Har bir juftlik yigʻindisini hisoblang.

$$-1-66=-67 -2-33=-35 -3-22=-25 -6-11=-17$$

Yechim – -17 yigʻindisini beruvchi juftlik.

$$a=-6 b=-11$$

-3x^{2}-17x-22 ni \left(-3x^{2}-6x\right)+\left(-11x-22\right) sifatida qaytadan yozish.

$$\left(-3x^{2}-6x\right)+\left(-11x-22\right)$$

Birinchi guruhda 3x ni va ikkinchi guruhda 11 ni faktordan chiqaring.

$$3x\left(-x-2\right)+11\left(-x-2\right)$$

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

$$\left(-x-2\right)\left(3x+11\right)$$

Tenglamani yechish uchun -x-2=0 va 3x+11=0 ni yeching.

$$x=-2 x=-\frac{11}{3}$$

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.

$$-3x^{2}-17x=22$$

Tenglamaning ikkala tarafidan 22 ni ayirish.

$$-3x^{2}-17x-22=22-22$$

O‘zidan 22 ayirilsa 0 qoladi.

$$-3x^{2}-17x-22=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} -3 ni a, -17 ni b va -22 ni c bilan almashtiring.

$$x=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\left(-3\right)\left(-22\right)}}{2\left(-3\right)}$$

-17 kvadratini chiqarish.

$$x=\frac{-\left(-17\right)±\sqrt{289-4\left(-3\right)\left(-22\right)}}{2\left(-3\right)}$$

-4 ni -3 marotabaga ko'paytirish.

$$x=\frac{-\left(-17\right)±\sqrt{289+12\left(-22\right)}}{2\left(-3\right)}$$

12 ni -22 marotabaga ko'paytirish.

$$x=\frac{-\left(-17\right)±\sqrt{289-264}}{2\left(-3\right)}$$

289 ni -264 ga qo'shish.

$$x=\frac{-\left(-17\right)±\sqrt{25}}{2\left(-3\right)}$$

25 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-17\right)±5}{2\left(-3\right)}$$

-17 ning teskarisi 17 ga teng.

$$x=\frac{17±5}{2\left(-3\right)}$$

2 ni -3 marotabaga ko'paytirish.

$$x=\frac{17±5}{-6}$$

x=\frac{17±5}{-6} tenglamasini yeching, bunda ± musbat. 17 ni 5 ga qo'shish.

$$x=\frac{22}{-6}$$

\frac{22}{-6} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

x=\frac{17±5}{-6} tenglamasini yeching, bunda ± manfiy. 17 dan 5 ni ayirish.

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

12 ni -6 ga bo'lish.

$$x=-2$$

Tenglama yechildi.

$$x=-\frac{11}{3} x=-2$$

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.

$$-3x^{2}-17x=22$$

Ikki tarafini -3 ga bo‘ling.

$$\frac{-3x^{2}-17x}{-3}=\frac{22}{-3}$$

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

$$x^{2}+\left(-\frac{17}{-3}\right)x=\frac{22}{-3}$$

-17 ni -3 ga bo'lish.

$$x^{2}+\frac{17}{3}x=\frac{22}{-3}$$

22 ni -3 ga bo'lish.

$$x^{2}+\frac{17}{3}x=-\frac{22}{3}$$

\frac{17}{3} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{17}{6} olish uchun. Keyin, \frac{17}{6} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{17}{3}x+\left(\frac{17}{6}\right)^{2}=-\frac{22}{3}+\left(\frac{17}{6}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{17}{6} kvadratini chiqarish.

$$x^{2}+\frac{17}{3}x+\frac{289}{36}=-\frac{22}{3}+\frac{289}{36}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{22}{3} ni \frac{289}{36} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{17}{3}x+\frac{289}{36}=\frac{25}{36}$$

x^{2}+\frac{17}{3}x+\frac{289}{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.

$$\left(x+\frac{17}{6}\right)^{2}=\frac{25}{36}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{17}{6}\right)^{2}}=\sqrt{\frac{25}{36}}$$

Qisqartirish.

$$x+\frac{17}{6}=\frac{5}{6} x+\frac{17}{6}=-\frac{5}{6}$$

Tenglamaning ikkala tarafidan \frac{17}{6} ni ayirish.

$$x=-2 x=-\frac{11}{3}$$