2x*3+x/5=5 eching

$$2x \times \frac{ 3+x }{ 5 } =5$$

x uchun yechish

$$x = \frac{\sqrt{59} - 3}{2} \approx 2,340572874$$

$$x=\frac{-\sqrt{59}-3}{2}\approx -5,340572874$$

Kvadrat tenglamalaridan foydalanish qoidalari

Tenglamaning ikkala tarafini 5 ga ko'paytirish.

$$2x\left(3+x\right)=25$$

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

$$6x+2x^{2}=25$$

Ikkala tarafdan 25 ni ayirish.

$$6x+2x^{2}-25=0$$

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.

$$2x^{2}+6x-25=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 2 ni a, 6 ni b va -25 ni c bilan almashtiring.

$$x=\frac{-6±\sqrt{6^{2}-4\times 2\left(-25\right)}}{2\times 2}$$

6 kvadratini chiqarish.

$$x=\frac{-6±\sqrt{36-4\times 2\left(-25\right)}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36-8\left(-25\right)}}{2\times 2}$$

-8 ni -25 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36+200}}{2\times 2}$$

36 ni 200 ga qo'shish.

$$x=\frac{-6±\sqrt{236}}{2\times 2}$$

236 ning kvadrat ildizini chiqarish.

$$x=\frac{-6±2\sqrt{59}}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{-6±2\sqrt{59}}{4}$$

x=\frac{-6±2\sqrt{59}}{4} tenglamasini yeching, bunda ± musbat. -6 ni 2\sqrt{59} ga qo'shish.

$$x=\frac{2\sqrt{59}-6}{4}$$

-6+2\sqrt{59} ni 4 ga bo'lish.

$$x=\frac{\sqrt{59}-3}{2}$$

x=\frac{-6±2\sqrt{59}}{4} tenglamasini yeching, bunda ± manfiy. -6 dan 2\sqrt{59} ni ayirish.

$$x=\frac{-2\sqrt{59}-6}{4}$$

-6-2\sqrt{59} ni 4 ga bo'lish.

$$x=\frac{-\sqrt{59}-3}{2}$$

Tenglama yechildi.

$$x=\frac{\sqrt{59}-3}{2} x=\frac{-\sqrt{59}-3}{2}$$

Kvadratni yakunlash bo'yicha qadamlar

Tenglamaning ikkala tarafini 5 ga ko'paytirish.

$$2x\left(3+x\right)=25$$

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

$$6x+2x^{2}=25$$

Bu kabi kvadrat tenglamalarni kvadratni yakunlab yechish mumkin. Kvadratni yechish uchun tenglama avval ushbu shaklda bo'lishi shart: x^{2}+bx=c.

$$2x^{2}+6x=25$$

Ikki tarafini 2 ga bo‘ling.

$$\frac{2x^{2}+6x}{2}=\frac{25}{2}$$

2 ga bo'lish 2 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{6}{2}x=\frac{25}{2}$$

6 ni 2 ga bo'lish.

$$x^{2}+3x=\frac{25}{2}$$

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.

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=\frac{25}{2}+\left(\frac{3}{2}\right)^{2}$$

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

$$x^{2}+3x+\frac{9}{4}=\frac{25}{2}+\frac{9}{4}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{25}{2} ni \frac{9}{4} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+3x+\frac{9}{4}=\frac{59}{4}$$

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.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{59}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{59}{4}}$$

Qisqartirish.

$$x+\frac{3}{2}=\frac{\sqrt{59}}{2} x+\frac{3}{2}=-\frac{\sqrt{59}}{2}$$

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

$$x=\frac{\sqrt{59}-3}{2} x=\frac{-\sqrt{59}-3}{2}$$