x^2+x-1=0 eching

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

x uchun yechish

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

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

Kvadrat tenglamalaridan foydalanish qoidalari

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.

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

1 kvadratini chiqarish.

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

-4 ni -1 marotabaga ko'paytirish.

$$x=\frac{-1±\sqrt{1+4}}{2}$$

1 ni 4 ga qo'shish.

$$x=\frac{-1±\sqrt{5}}{2}$$

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

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

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

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

Tenglama yechildi.

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

Kvadratni yakunlash bo'yicha qadamlar

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

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

1 ni tenglamaning ikkala tarafiga qo'shish.

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

O‘zidan -1 ayirilsa 0 qoladi.

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

0 dan -1 ni ayirish.

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

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.

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

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

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