2x^2+14x-4=-x^2+3x eching

$$2 x ^ { 2 } + 14 x - 4 = - x ^ { 2 } + 3 x$$

x uchun yechish

$$x=-4$$

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

Guruhlash orqali faktorlash qadamlari

x^{2} ni ikki tarafga qo’shing.

$$2x^{2}+14x-4+x^{2}=3x$$

3x^{2} ni olish uchun 2x^{2} va x^{2} ni birlashtirish.

$$3x^{2}+14x-4=3x$$

Ikkala tarafdan 3x ni ayirish.

$$3x^{2}+14x-4-3x=0$$

11x ni olish uchun 14x va -3x ni birlashtirish.

$$3x^{2}+11x-4=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon 3x^{2}+ax+bx-4 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=11 ab=3\left(-4\right)=-12$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b musbat boʻlganda, musbat sonda manfiyga nisbatdan kattaroq mutlaq qiymat bor. -12-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,12 -2,6 -3,4$$

Har bir juftlik yigʻindisini hisoblang.

$$-1+12=11 -2+6=4 -3+4=1$$

Yechim – 11 yigʻindisini beruvchi juftlik.

$$a=-1 b=12$$

3x^{2}+11x-4 ni \left(3x^{2}-x\right)+\left(12x-4\right) sifatida qaytadan yozish.

$$\left(3x^{2}-x\right)+\left(12x-4\right)$$

Birinchi guruhda x ni va ikkinchi guruhda 4 ni faktordan chiqaring.

$$x\left(3x-1\right)+4\left(3x-1\right)$$

Distributiv funktsiyasidan foydalangan holda 3x-1 umumiy terminini chiqaring.

$$\left(3x-1\right)\left(x+4\right)$$

Tenglamani yechish uchun 3x-1=0 va x+4=0 ni yeching.

$$x=\frac{1}{3} x=-4$$

Kvadrat tenglamalaridan foydalanish qoidalari

x^{2} ni ikki tarafga qo’shing.

$$2x^{2}+14x-4+x^{2}=3x$$

3x^{2} ni olish uchun 2x^{2} va x^{2} ni birlashtirish.

$$3x^{2}+14x-4=3x$$

Ikkala tarafdan 3x ni ayirish.

$$3x^{2}+14x-4-3x=0$$

11x ni olish uchun 14x va -3x ni birlashtirish.

$$3x^{2}+11x-4=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 3 ni a, 11 ni b va -4 ni c bilan almashtiring.

$$x=\frac{-11±\sqrt{11^{2}-4\times 3\left(-4\right)}}{2\times 3}$$

11 kvadratini chiqarish.

$$x=\frac{-11±\sqrt{121-4\times 3\left(-4\right)}}{2\times 3}$$

-4 ni 3 marotabaga ko'paytirish.

$$x=\frac{-11±\sqrt{121-12\left(-4\right)}}{2\times 3}$$

-12 ni -4 marotabaga ko'paytirish.

$$x=\frac{-11±\sqrt{121+48}}{2\times 3}$$

121 ni 48 ga qo'shish.

$$x=\frac{-11±\sqrt{169}}{2\times 3}$$

169 ning kvadrat ildizini chiqarish.

$$x=\frac{-11±13}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$x=\frac{-11±13}{6}$$

x=\frac{-11±13}{6} tenglamasini yeching, bunda ± musbat. -11 ni 13 ga qo'shish.

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

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

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

x=\frac{-11±13}{6} tenglamasini yeching, bunda ± manfiy. -11 dan 13 ni ayirish.

$$x=-\frac{24}{6}$$

-24 ni 6 ga bo'lish.

$$x=-4$$

Tenglama yechildi.

$$x=\frac{1}{3} x=-4$$

Kvadratni yakunlash bo'yicha qadamlar

x^{2} ni ikki tarafga qo’shing.

$$2x^{2}+14x-4+x^{2}=3x$$

3x^{2} ni olish uchun 2x^{2} va x^{2} ni birlashtirish.

$$3x^{2}+14x-4=3x$$

Ikkala tarafdan 3x ni ayirish.

$$3x^{2}+14x-4-3x=0$$

11x ni olish uchun 14x va -3x ni birlashtirish.

$$3x^{2}+11x-4=0$$

4 ni ikki tarafga qo’shing. Har qanday songa nolni qo‘shsangiz, o‘zi chiqadi.

$$3x^{2}+11x=4$$

Ikki tarafini 3 ga bo‘ling.

$$\frac{3x^{2}+11x}{3}=\frac{4}{3}$$

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

$$x^{2}+\frac{11}{3}x=\frac{4}{3}$$

\frac{11}{3} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{11}{6} olish uchun. Keyin, \frac{11}{6} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{11}{3}x+\left(\frac{11}{6}\right)^{2}=\frac{4}{3}+\left(\frac{11}{6}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{11}{6} kvadratini chiqarish.

$$x^{2}+\frac{11}{3}x+\frac{121}{36}=\frac{4}{3}+\frac{121}{36}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{4}{3} ni \frac{121}{36} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{11}{3}x+\frac{121}{36}=\frac{169}{36}$$

x^{2}+\frac{11}{3}x+\frac{121}{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.

$$\left(x+\frac{11}{6}\right)^{2}=\frac{169}{36}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{11}{6}\right)^{2}}=\sqrt{\frac{169}{36}}$$

Qisqartirish.

$$x+\frac{11}{6}=\frac{13}{6} x+\frac{11}{6}=-\frac{13}{6}$$

Tenglamaning ikkala tarafidan \frac{11}{6} ni ayirish.

$$x=\frac{1}{3} x=-4$$