3x^2-5x-250=0 eching

$$3 { x }^{ 2 } -5x-250=0$$

x uchun yechish

$$x = -\frac{25}{3} = -8\frac{1}{3} \approx -8,333333333$$

$$x=10$$

Guruhlash orqali faktorlash qadamlari

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon 3x^{2}+ax+bx-250 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-5 ab=3\left(-250\right)=-750$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -750-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-750 2,-375 3,-250 5,-150 6,-125 10,-75 15,-50 25,-30$$

Har bir juftlik yigʻindisini hisoblang.

$$1-750=-749 2-375=-373 3-250=-247 5-150=-145 6-125=-119 10-75=-65 15-50=-35 25-30=-5$$

Yechim – -5 yigʻindisini beruvchi juftlik.

$$a=-30 b=25$$

3x^{2}-5x-250 ni \left(3x^{2}-30x\right)+\left(25x-250\right) sifatida qaytadan yozish.

$$\left(3x^{2}-30x\right)+\left(25x-250\right)$$

Birinchi guruhda 3x ni va ikkinchi guruhda 25 ni faktordan chiqaring.

$$3x\left(x-10\right)+25\left(x-10\right)$$

Distributiv funktsiyasidan foydalangan holda x-10 umumiy terminini chiqaring.

$$\left(x-10\right)\left(3x+25\right)$$

Tenglamani yechish uchun x-10=0 va 3x+25=0 ni yeching.

$$x=10 x=-\frac{25}{3}$$

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.

$$3x^{2}-5x-250=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 3 ni a, -5 ni b va -250 ni c bilan almashtiring.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 3\left(-250\right)}}{2\times 3}$$

-5 kvadratini chiqarish.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 3\left(-250\right)}}{2\times 3}$$

-4 ni 3 marotabaga ko'paytirish.

$$x=\frac{-\left(-5\right)±\sqrt{25-12\left(-250\right)}}{2\times 3}$$

-12 ni -250 marotabaga ko'paytirish.

$$x=\frac{-\left(-5\right)±\sqrt{25+3000}}{2\times 3}$$

25 ni 3000 ga qo'shish.

$$x=\frac{-\left(-5\right)±\sqrt{3025}}{2\times 3}$$

3025 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-5\right)±55}{2\times 3}$$

-5 ning teskarisi 5 ga teng.

$$x=\frac{5±55}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$x=\frac{5±55}{6}$$

x=\frac{5±55}{6} tenglamasini yeching, bunda ± musbat. 5 ni 55 ga qo'shish.

$$x=\frac{60}{6}$$

60 ni 6 ga bo'lish.

$$x=10$$

x=\frac{5±55}{6} tenglamasini yeching, bunda ± manfiy. 5 dan 55 ni ayirish.

$$x=-\frac{50}{6}$$

\frac{-50}{6} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=-\frac{25}{3}$$

Tenglama yechildi.

$$x=10 x=-\frac{25}{3}$$

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.

$$3x^{2}-5x-250=0$$

250 ni tenglamaning ikkala tarafiga qo'shish.

$$3x^{2}-5x-250-\left(-250\right)=-\left(-250\right)$$

O‘zidan -250 ayirilsa 0 qoladi.

$$3x^{2}-5x=-\left(-250\right)$$

0 dan -250 ni ayirish.

$$3x^{2}-5x=250$$

Ikki tarafini 3 ga bo‘ling.

$$\frac{3x^{2}-5x}{3}=\frac{250}{3}$$

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

$$x^{2}-\frac{5}{3}x=\frac{250}{3}$$

-\frac{5}{3} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{5}{6} olish uchun. Keyin, -\frac{5}{6} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-\frac{5}{3}x+\left(-\frac{5}{6}\right)^{2}=\frac{250}{3}+\left(-\frac{5}{6}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{5}{6} kvadratini chiqarish.

$$x^{2}-\frac{5}{3}x+\frac{25}{36}=\frac{250}{3}+\frac{25}{36}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{250}{3} ni \frac{25}{36} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}-\frac{5}{3}x+\frac{25}{36}=\frac{3025}{36}$$

x^{2}-\frac{5}{3}x+\frac{25}{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.

$$\left(x-\frac{5}{6}\right)^{2}=\frac{3025}{36}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{5}{6}\right)^{2}}=\sqrt{\frac{3025}{36}}$$

Qisqartirish.

$$x-\frac{5}{6}=\frac{55}{6} x-\frac{5}{6}=-\frac{55}{6}$$

\frac{5}{6} ni tenglamaning ikkala tarafiga qo'shish.

$$x=10 x=-\frac{25}{3}$$