12t^2+17t=40 eching

$$12 { t }^{ 2 } +17t = 40$$

t uchun yechish

$$t = -\frac{8}{3} = -2\frac{2}{3} \approx -2,666666667$$

$$t = \frac{5}{4} = 1\frac{1}{4} = 1,25$$

Guruhlash orqali faktorlash qadamlari

Ikkala tarafdan 40 ni ayirish.

$$12t^{2}+17t-40=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon 12t^{2}+at+bt-40 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=17 ab=12\left(-40\right)=-480$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b musbat boʻlganda, musbat sonda manfiyga nisbatdan kattaroq mutlaq qiymat bor. -480-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,480 -2,240 -3,160 -4,120 -5,96 -6,80 -8,60 -10,48 -12,40 -15,32 -16,30 -20,24$$

Har bir juftlik yigʻindisini hisoblang.

$$-1+480=479 -2+240=238 -3+160=157 -4+120=116 -5+96=91 -6+80=74 -8+60=52 -10+48=38 -12+40=28 -15+32=17 -16+30=14 -20+24=4$$

Yechim – 17 yigʻindisini beruvchi juftlik.

$$a=-15 b=32$$

12t^{2}+17t-40 ni \left(12t^{2}-15t\right)+\left(32t-40\right) sifatida qaytadan yozish.

$$\left(12t^{2}-15t\right)+\left(32t-40\right)$$

Birinchi guruhda 3t ni va ikkinchi guruhda 8 ni faktordan chiqaring.

$$3t\left(4t-5\right)+8\left(4t-5\right)$$

Distributiv funktsiyasidan foydalangan holda 4t-5 umumiy terminini chiqaring.

$$\left(4t-5\right)\left(3t+8\right)$$

Tenglamani yechish uchun 4t-5=0 va 3t+8=0 ni yeching.

$$t=\frac{5}{4} t=-\frac{8}{3}$$

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.

$$12t^{2}+17t=40$$

Tenglamaning ikkala tarafidan 40 ni ayirish.

$$12t^{2}+17t-40=40-40$$

O‘zidan 40 ayirilsa 0 qoladi.

$$12t^{2}+17t-40=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 12 ni a, 17 ni b va -40 ni c bilan almashtiring.

$$t=\frac{-17±\sqrt{17^{2}-4\times 12\left(-40\right)}}{2\times 12}$$

17 kvadratini chiqarish.

$$t=\frac{-17±\sqrt{289-4\times 12\left(-40\right)}}{2\times 12}$$

-4 ni 12 marotabaga ko'paytirish.

$$t=\frac{-17±\sqrt{289-48\left(-40\right)}}{2\times 12}$$

-48 ni -40 marotabaga ko'paytirish.

$$t=\frac{-17±\sqrt{289+1920}}{2\times 12}$$

289 ni 1920 ga qo'shish.

$$t=\frac{-17±\sqrt{2209}}{2\times 12}$$

2209 ning kvadrat ildizini chiqarish.

$$t=\frac{-17±47}{2\times 12}$$

2 ni 12 marotabaga ko'paytirish.

$$t=\frac{-17±47}{24}$$

t=\frac{-17±47}{24} tenglamasini yeching, bunda ± musbat. -17 ni 47 ga qo'shish.

$$t=\frac{30}{24}$$

\frac{30}{24} ulushini 6 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$t=\frac{5}{4}$$

t=\frac{-17±47}{24} tenglamasini yeching, bunda ± manfiy. -17 dan 47 ni ayirish.

$$t=-\frac{64}{24}$$

\frac{-64}{24} ulushini 8 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$t=-\frac{8}{3}$$

Tenglama yechildi.

$$t=\frac{5}{4} t=-\frac{8}{3}$$

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.

$$12t^{2}+17t=40$$

Ikki tarafini 12 ga bo‘ling.

$$\frac{12t^{2}+17t}{12}=\frac{40}{12}$$

12 ga bo'lish 12 ga ko'paytirishni bekor qiladi.

$$t^{2}+\frac{17}{12}t=\frac{40}{12}$$

\frac{40}{12} ulushini 4 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$t^{2}+\frac{17}{12}t=\frac{10}{3}$$

\frac{17}{12} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{17}{24} olish uchun. Keyin, \frac{17}{24} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$t^{2}+\frac{17}{12}t+\left(\frac{17}{24}\right)^{2}=\frac{10}{3}+\left(\frac{17}{24}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{17}{24} kvadratini chiqarish.

$$t^{2}+\frac{17}{12}t+\frac{289}{576}=\frac{10}{3}+\frac{289}{576}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{10}{3} ni \frac{289}{576} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$t^{2}+\frac{17}{12}t+\frac{289}{576}=\frac{2209}{576}$$

t^{2}+\frac{17}{12}t+\frac{289}{576} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(t+\frac{17}{24}\right)^{2}=\frac{2209}{576}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(t+\frac{17}{24}\right)^{2}}=\sqrt{\frac{2209}{576}}$$

Qisqartirish.

$$t+\frac{17}{24}=\frac{47}{24} t+\frac{17}{24}=-\frac{47}{24}$$

Tenglamaning ikkala tarafidan \frac{17}{24} ni ayirish.

$$t=\frac{5}{4} t=-\frac{8}{3}$$