lambda^2+(1-lambda)^2=3 eching

$$\lambda ^ { 2 } + ( 1 - \lambda ) ^ { 2 } = 3$$

λ uchun yechish

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

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

Kvadrat tenglamalaridan foydalanish qoidalari

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(1-\lambda \right)^{2} kengaytirilishi uchun ishlating.

$$\lambda ^{2}+1-2\lambda +\lambda ^{2}=3$$

2\lambda ^{2} ni olish uchun \lambda ^{2} va \lambda ^{2} ni birlashtirish.

$$2\lambda ^{2}+1-2\lambda =3$$

Ikkala tarafdan 3 ni ayirish.

$$2\lambda ^{2}+1-2\lambda -3=0$$

-2 olish uchun 1 dan 3 ni ayirish.

$$2\lambda ^{2}-2-2\lambda =0$$

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.

$$2\lambda ^{2}-2\lambda -2=0$$

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

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

-2 kvadratini chiqarish.

$$\lambda =\frac{-\left(-2\right)±\sqrt{4-4\times 2\left(-2\right)}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$\lambda =\frac{-\left(-2\right)±\sqrt{4-8\left(-2\right)}}{2\times 2}$$

-8 ni -2 marotabaga ko'paytirish.

$$\lambda =\frac{-\left(-2\right)±\sqrt{4+16}}{2\times 2}$$

4 ni 16 ga qo'shish.

$$\lambda =\frac{-\left(-2\right)±\sqrt{20}}{2\times 2}$$

20 ning kvadrat ildizini chiqarish.

$$\lambda =\frac{-\left(-2\right)±2\sqrt{5}}{2\times 2}$$

-2 ning teskarisi 2 ga teng.

$$\lambda =\frac{2±2\sqrt{5}}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$\lambda =\frac{2±2\sqrt{5}}{4}$$

\lambda =\frac{2±2\sqrt{5}}{4} tenglamasini yeching, bunda ± musbat. 2 ni 2\sqrt{5} ga qo'shish.

$$\lambda =\frac{2\sqrt{5}+2}{4}$$

2+2\sqrt{5} ni 4 ga bo'lish.

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

\lambda =\frac{2±2\sqrt{5}}{4} tenglamasini yeching, bunda ± manfiy. 2 dan 2\sqrt{5} ni ayirish.

$$\lambda =\frac{2-2\sqrt{5}}{4}$$

2-2\sqrt{5} ni 4 ga bo'lish.

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

Tenglama yechildi.

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

Kvadratni yakunlash bo'yicha qadamlar

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(1-\lambda \right)^{2} kengaytirilishi uchun ishlating.

$$\lambda ^{2}+1-2\lambda +\lambda ^{2}=3$$

2\lambda ^{2} ni olish uchun \lambda ^{2} va \lambda ^{2} ni birlashtirish.

$$2\lambda ^{2}+1-2\lambda =3$$

Ikkala tarafdan 1 ni ayirish.

$$2\lambda ^{2}-2\lambda =3-1$$

2 olish uchun 3 dan 1 ni ayirish.

$$2\lambda ^{2}-2\lambda =2$$

Ikki tarafini 2 ga bo‘ling.

$$\frac{2\lambda ^{2}-2\lambda }{2}=\frac{2}{2}$$

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

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

-2 ni 2 ga bo'lish.

$$\lambda ^{2}-\lambda =\frac{2}{2}$$

2 ni 2 ga bo'lish.

$$\lambda ^{2}-\lambda =1$$

-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.

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

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

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

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

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

\lambda ^{2}-\lambda +\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.

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

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

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

Qisqartirish.

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

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

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