x=x^2-1 eching

$$x = x ^ { 2 } - 1$$

x uchun yechish

$$x = \frac{\sqrt{5} + 1}{2} \approx 1,618033989$$

$$x=\frac{1-\sqrt{5}}{2}\approx -0,618033989$$

Kvadrat tenglamalaridan foydalanish qoidalari

Ikkala tarafdan x^{2} ni ayirish.

$$x-x^{2}=-1$$

1 ni ikki tarafga qo’shing.

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

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.

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

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 1 ni c bilan almashtiring.

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

1 kvadratini chiqarish.

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

-4 ni -1 marotabaga ko'paytirish.

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

1 ni 4 ga qo'shish.

$$x=\frac{-1±\sqrt{5}}{2\left(-1\right)}$$

2 ni -1 marotabaga ko'paytirish.

$$x=\frac{-1±\sqrt{5}}{-2}$$

x=\frac{-1±\sqrt{5}}{-2} tenglamasini yeching, bunda ± musbat. -1 ni \sqrt{5} ga qo'shish.

$$x=\frac{\sqrt{5}-1}{-2}$$

-1+\sqrt{5} ni -2 ga bo'lish.

$$x=\frac{1-\sqrt{5}}{2}$$

x=\frac{-1±\sqrt{5}}{-2} tenglamasini yeching, bunda ± manfiy. -1 dan \sqrt{5} ni ayirish.

$$x=\frac{-\sqrt{5}-1}{-2}$$

-1-\sqrt{5} ni -2 ga bo'lish.

$$x=\frac{\sqrt{5}+1}{2}$$

Tenglama yechildi.

$$x=\frac{1-\sqrt{5}}{2} x=\frac{\sqrt{5}+1}{2}$$

Kvadratni yakunlash bo'yicha qadamlar

Ikkala tarafdan x^{2} ni ayirish.

$$x-x^{2}=-1$$

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

$$-x^{2}+x=-1$$

Ikki tarafini -1 ga bo‘ling.

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

-1 ga bo'lish -1 ga ko'paytirishni bekor qiladi.

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

1 ni -1 ga bo'lish.

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

-1 ni -1 ga bo'lish.

$$x^{2}-x=1$$

-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.

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

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

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

1 ni \frac{1}{4} ga qo'shish.

$$x^{2}-x+\frac{1}{4}=\frac{5}{4}$$

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.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{5}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{5}{4}}$$

Qisqartirish.

$$x-\frac{1}{2}=\frac{\sqrt{5}}{2} x-\frac{1}{2}=-\frac{\sqrt{5}}{2}$$

\frac{1}{2} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{\sqrt{5}+1}{2} x=\frac{1-\sqrt{5}}{2}$$