x(x+2)+x=2 eching

$$x(x+2)+x=2$$

x uchun yechish

$$x=\frac{\sqrt{17}-3}{2}\approx 0,561552813$$

$$x=\frac{-\sqrt{17}-3}{2}\approx -3,561552813$$

Kvadrat tenglamalaridan foydalanish qoidalari

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

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

3x ni olish uchun 2x va x ni birlashtirish.

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

Ikkala tarafdan 2 ni ayirish.

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

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

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

3 kvadratini chiqarish.

$$x=\frac{-3±\sqrt{9-4\left(-2\right)}}{2}$$

-4 ni -2 marotabaga ko'paytirish.

$$x=\frac{-3±\sqrt{9+8}}{2}$$

9 ni 8 ga qo'shish.

$$x=\frac{-3±\sqrt{17}}{2}$$

x=\frac{-3±\sqrt{17}}{2} tenglamasini yeching, bunda ± musbat. -3 ni \sqrt{17} ga qo'shish.

$$x=\frac{\sqrt{17}-3}{2}$$

x=\frac{-3±\sqrt{17}}{2} tenglamasini yeching, bunda ± manfiy. -3 dan \sqrt{17} ni ayirish.

$$x=\frac{-\sqrt{17}-3}{2}$$

Tenglama yechildi.

$$x=\frac{\sqrt{17}-3}{2} x=\frac{-\sqrt{17}-3}{2}$$

Kvadratni yakunlash bo'yicha qadamlar

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

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

3x ni olish uchun 2x va x ni birlashtirish.

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

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.

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

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

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

2 ni \frac{9}{4} ga qo'shish.

$$x^{2}+3x+\frac{9}{4}=\frac{17}{4}$$

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.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{17}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{17}{4}}$$

Qisqartirish.

$$x+\frac{3}{2}=\frac{\sqrt{17}}{2} x+\frac{3}{2}=-\frac{\sqrt{17}}{2}$$

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

$$x=\frac{\sqrt{17}-3}{2} x=\frac{-\sqrt{17}-3}{2}$$