1/2=x(x+1) eching

$$\frac{ 1 }{ 2 } =x(x+1)$$

x uchun yechish

$$x=\frac{\sqrt{3}-1}{2}\approx 0,366025404$$

$$x=\frac{-\sqrt{3}-1}{2}\approx -1,366025404$$

Kvadrat tenglamalaridan foydalanish qoidalari

x ga x+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

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

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

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

Ikkala tarafdan \frac{1}{2} ni ayirish.

$$x^{2}+x-\frac{1}{2}=0$$

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

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

1 kvadratini chiqarish.

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

-4 ni -\frac{1}{2} marotabaga ko'paytirish.

$$x=\frac{-1±\sqrt{1+2}}{2}$$

1 ni 2 ga qo'shish.

$$x=\frac{-1±\sqrt{3}}{2}$$

x=\frac{-1±\sqrt{3}}{2} tenglamasini yeching, bunda ± musbat. -1 ni \sqrt{3} ga qo'shish.

$$x=\frac{\sqrt{3}-1}{2}$$

x=\frac{-1±\sqrt{3}}{2} tenglamasini yeching, bunda ± manfiy. -1 dan \sqrt{3} ni ayirish.

$$x=\frac{-\sqrt{3}-1}{2}$$

Tenglama yechildi.

$$x=\frac{\sqrt{3}-1}{2} x=\frac{-\sqrt{3}-1}{2}$$

Kvadratni yakunlash bo'yicha qadamlar

x ga x+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

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

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

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

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

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{1}{2} kvadratini chiqarish.

$$x^{2}+x+\frac{1}{4}=\frac{1}{2}+\frac{1}{4}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{1}{2} ni \frac{1}{4} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+x+\frac{1}{4}=\frac{3}{4}$$

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

$$\left(x+\frac{1}{2}\right)^{2}=\frac{3}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{3}{4}}$$

Qisqartirish.

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

Tenglamaning ikkala tarafidan \frac{1}{2} ni ayirish.

$$x=\frac{\sqrt{3}-1}{2} x=\frac{-\sqrt{3}-1}{2}$$