3x(x+2)=5(x+2) eching

$$3x(x+2)=5(x+2)$$

x uchun yechish

$$x=-2$$

$$x = \frac{5}{3} = 1\frac{2}{3} \approx 1,666666667$$

Kvadrat tenglamalaridan foydalanish qoidalari

3x ga x+2 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

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

5 ga x+2 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$3x^{2}+6x=5x+10$$

Ikkala tarafdan 5x ni ayirish.

$$3x^{2}+6x-5x=10$$

x ni olish uchun 6x va -5x ni birlashtirish.

$$3x^{2}+x=10$$

Ikkala tarafdan 10 ni ayirish.

$$3x^{2}+x-10=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 3 ni a, 1 ni b va -10 ni c bilan almashtiring.

$$x=\frac{-1±\sqrt{1^{2}-4\times 3\left(-10\right)}}{2\times 3}$$

1 kvadratini chiqarish.

$$x=\frac{-1±\sqrt{1-4\times 3\left(-10\right)}}{2\times 3}$$

-4 ni 3 marotabaga ko'paytirish.

$$x=\frac{-1±\sqrt{1-12\left(-10\right)}}{2\times 3}$$

-12 ni -10 marotabaga ko'paytirish.

$$x=\frac{-1±\sqrt{1+120}}{2\times 3}$$

1 ni 120 ga qo'shish.

$$x=\frac{-1±\sqrt{121}}{2\times 3}$$

121 ning kvadrat ildizini chiqarish.

$$x=\frac{-1±11}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$x=\frac{-1±11}{6}$$

x=\frac{-1±11}{6} tenglamasini yeching, bunda ± musbat. -1 ni 11 ga qo'shish.

$$x=\frac{10}{6}$$

\frac{10}{6} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=\frac{5}{3}$$

x=\frac{-1±11}{6} tenglamasini yeching, bunda ± manfiy. -1 dan 11 ni ayirish.

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

-12 ni 6 ga bo'lish.

$$x=-2$$

Tenglama yechildi.

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

Kvadratni yakunlash bo'yicha qadamlar

3x ga x+2 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

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

5 ga x+2 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$3x^{2}+6x=5x+10$$

Ikkala tarafdan 5x ni ayirish.

$$3x^{2}+6x-5x=10$$

x ni olish uchun 6x va -5x ni birlashtirish.

$$3x^{2}+x=10$$

Ikki tarafini 3 ga bo‘ling.

$$\frac{3x^{2}+x}{3}=\frac{10}{3}$$

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

$$x^{2}+\frac{1}{3}x=\frac{10}{3}$$

\frac{1}{3} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{1}{6} olish uchun. Keyin, \frac{1}{6} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{1}{3}x+\left(\frac{1}{6}\right)^{2}=\frac{10}{3}+\left(\frac{1}{6}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{1}{6} kvadratini chiqarish.

$$x^{2}+\frac{1}{3}x+\frac{1}{36}=\frac{10}{3}+\frac{1}{36}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{10}{3} ni \frac{1}{36} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{1}{3}x+\frac{1}{36}=\frac{121}{36}$$

x^{2}+\frac{1}{3}x+\frac{1}{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.

$$\left(x+\frac{1}{6}\right)^{2}=\frac{121}{36}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{1}{6}\right)^{2}}=\sqrt{\frac{121}{36}}$$

Qisqartirish.

$$x+\frac{1}{6}=\frac{11}{6} x+\frac{1}{6}=-\frac{11}{6}$$

Tenglamaning ikkala tarafidan \frac{1}{6} ni ayirish.

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