3x^2+53x-720=0 eching

$$3 { x }^{ 2 } +53x-720=0$$

x uchun yechish

$$x = -\frac{80}{3} = -26\frac{2}{3} \approx -26,666666667$$

$$x=9$$

Guruhlash orqali faktorlash qadamlari

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon 3x^{2}+ax+bx-720 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=53 ab=3\left(-720\right)=-2160$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b musbat boʻlganda, musbat sonda manfiyga nisbatdan kattaroq mutlaq qiymat bor. -2160-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,2160 -2,1080 -3,720 -4,540 -5,432 -6,360 -8,270 -9,240 -10,216 -12,180 -15,144 -16,135 -18,120 -20,108 -24,90 -27,80 -30,72 -36,60 -40,54 -45,48$$

Har bir juftlik yigʻindisini hisoblang.

$$-1+2160=2159 -2+1080=1078 -3+720=717 -4+540=536 -5+432=427 -6+360=354 -8+270=262 -9+240=231 -10+216=206 -12+180=168 -15+144=129 -16+135=119 -18+120=102 -20+108=88 -24+90=66 -27+80=53 -30+72=42 -36+60=24 -40+54=14 -45+48=3$$

Yechim – 53 yigʻindisini beruvchi juftlik.

$$a=-27 b=80$$

3x^{2}+53x-720 ni \left(3x^{2}-27x\right)+\left(80x-720\right) sifatida qaytadan yozish.

$$\left(3x^{2}-27x\right)+\left(80x-720\right)$$

Birinchi guruhda 3x ni va ikkinchi guruhda 80 ni faktordan chiqaring.

$$3x\left(x-9\right)+80\left(x-9\right)$$

Distributiv funktsiyasidan foydalangan holda x-9 umumiy terminini chiqaring.

$$\left(x-9\right)\left(3x+80\right)$$

Tenglamani yechish uchun x-9=0 va 3x+80=0 ni yeching.

$$x=9 x=-\frac{80}{3}$$

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.

$$3x^{2}+53x-720=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 3 ni a, 53 ni b va -720 ni c bilan almashtiring.

$$x=\frac{-53±\sqrt{53^{2}-4\times 3\left(-720\right)}}{2\times 3}$$

53 kvadratini chiqarish.

$$x=\frac{-53±\sqrt{2809-4\times 3\left(-720\right)}}{2\times 3}$$

-4 ni 3 marotabaga ko'paytirish.

$$x=\frac{-53±\sqrt{2809-12\left(-720\right)}}{2\times 3}$$

-12 ni -720 marotabaga ko'paytirish.

$$x=\frac{-53±\sqrt{2809+8640}}{2\times 3}$$

2809 ni 8640 ga qo'shish.

$$x=\frac{-53±\sqrt{11449}}{2\times 3}$$

11449 ning kvadrat ildizini chiqarish.

$$x=\frac{-53±107}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$x=\frac{-53±107}{6}$$

x=\frac{-53±107}{6} tenglamasini yeching, bunda ± musbat. -53 ni 107 ga qo'shish.

$$x=\frac{54}{6}$$

54 ni 6 ga bo'lish.

$$x=9$$

x=\frac{-53±107}{6} tenglamasini yeching, bunda ± manfiy. -53 dan 107 ni ayirish.

$$x=-\frac{160}{6}$$

\frac{-160}{6} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=-\frac{80}{3}$$

Tenglama yechildi.

$$x=9 x=-\frac{80}{3}$$

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.

$$3x^{2}+53x-720=0$$

720 ni tenglamaning ikkala tarafiga qo'shish.

$$3x^{2}+53x-720-\left(-720\right)=-\left(-720\right)$$

O‘zidan -720 ayirilsa 0 qoladi.

$$3x^{2}+53x=-\left(-720\right)$$

0 dan -720 ni ayirish.

$$3x^{2}+53x=720$$

Ikki tarafini 3 ga bo‘ling.

$$\frac{3x^{2}+53x}{3}=\frac{720}{3}$$

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

$$x^{2}+\frac{53}{3}x=\frac{720}{3}$$

720 ni 3 ga bo'lish.

$$x^{2}+\frac{53}{3}x=240$$

\frac{53}{3} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{53}{6} olish uchun. Keyin, \frac{53}{6} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{53}{3}x+\left(\frac{53}{6}\right)^{2}=240+\left(\frac{53}{6}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{53}{6} kvadratini chiqarish.

$$x^{2}+\frac{53}{3}x+\frac{2809}{36}=240+\frac{2809}{36}$$

240 ni \frac{2809}{36} ga qo'shish.

$$x^{2}+\frac{53}{3}x+\frac{2809}{36}=\frac{11449}{36}$$

x^{2}+\frac{53}{3}x+\frac{2809}{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.

$$\left(x+\frac{53}{6}\right)^{2}=\frac{11449}{36}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{53}{6}\right)^{2}}=\sqrt{\frac{11449}{36}}$$

Qisqartirish.

$$x+\frac{53}{6}=\frac{107}{6} x+\frac{53}{6}=-\frac{107}{6}$$

Tenglamaning ikkala tarafidan \frac{53}{6} ni ayirish.

$$x=9 x=-\frac{80}{3}$$