5*(x+3)x=7 eching

$$5 \times (x+3)x=7$$

x uchun yechish

$$x=\frac{\sqrt{365}}{10}-\frac{3}{2}\approx 0,410497317$$

$$x=-\frac{\sqrt{365}}{10}-\frac{3}{2}\approx -3,410497317$$

Kvadrat tenglamalaridan foydalanish qoidalari

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

$$\left(5x+15\right)x=7$$

5x+15 ga x ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$5x^{2}+15x=7$$

Ikkala tarafdan 7 ni ayirish.

$$5x^{2}+15x-7=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 5 ni a, 15 ni b va -7 ni c bilan almashtiring.

$$x=\frac{-15±\sqrt{15^{2}-4\times 5\left(-7\right)}}{2\times 5}$$

15 kvadratini chiqarish.

$$x=\frac{-15±\sqrt{225-4\times 5\left(-7\right)}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-15±\sqrt{225-20\left(-7\right)}}{2\times 5}$$

-20 ni -7 marotabaga ko'paytirish.

$$x=\frac{-15±\sqrt{225+140}}{2\times 5}$$

225 ni 140 ga qo'shish.

$$x=\frac{-15±\sqrt{365}}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$x=\frac{-15±\sqrt{365}}{10}$$

x=\frac{-15±\sqrt{365}}{10} tenglamasini yeching, bunda ± musbat. -15 ni \sqrt{365} ga qo'shish.

$$x=\frac{\sqrt{365}-15}{10}$$

-15+\sqrt{365} ni 10 ga bo'lish.

$$x=\frac{\sqrt{365}}{10}-\frac{3}{2}$$

x=\frac{-15±\sqrt{365}}{10} tenglamasini yeching, bunda ± manfiy. -15 dan \sqrt{365} ni ayirish.

$$x=\frac{-\sqrt{365}-15}{10}$$

-15-\sqrt{365} ni 10 ga bo'lish.

$$x=-\frac{\sqrt{365}}{10}-\frac{3}{2}$$

Tenglama yechildi.

$$x=\frac{\sqrt{365}}{10}-\frac{3}{2} x=-\frac{\sqrt{365}}{10}-\frac{3}{2}$$

Kvadratni yakunlash bo'yicha qadamlar

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

$$\left(5x+15\right)x=7$$

5x+15 ga x ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$5x^{2}+15x=7$$

Ikki tarafini 5 ga bo‘ling.

$$\frac{5x^{2}+15x}{5}=\frac{7}{5}$$

5 ga bo'lish 5 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{15}{5}x=\frac{7}{5}$$

15 ni 5 ga bo'lish.

$$x^{2}+3x=\frac{7}{5}$$

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

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=\frac{7}{5}+\left(\frac{3}{2}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{3}{2} kvadratini chiqarish.

$$x^{2}+3x+\frac{9}{4}=\frac{7}{5}+\frac{9}{4}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{7}{5} ni \frac{9}{4} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+3x+\frac{9}{4}=\frac{73}{20}$$

x^{2}+3x+\frac{9}{4} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{73}{20}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{73}{20}}$$

Qisqartirish.

$$x+\frac{3}{2}=\frac{\sqrt{365}}{10} x+\frac{3}{2}=-\frac{\sqrt{365}}{10}$$

Tenglamaning ikkala tarafidan \frac{3}{2} ni ayirish.

$$x=\frac{\sqrt{365}}{10}-\frac{3}{2} x=-\frac{\sqrt{365}}{10}-\frac{3}{2}$$