x=left(x-6right)^2 eching

$$x= { \left(x-6 \right) }^{ 2 }$$

x uchun yechish

$$x=9$$

$$x=4$$

Guruhlash orqali faktorlash qadamlari

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(x-6\right)^{2} kengaytirilishi uchun ishlating.

$$x=x^{2}-12x+36$$

Ikkala tarafdan x^{2} ni ayirish.

$$x-x^{2}=-12x+36$$

12x ni ikki tarafga qo’shing.

$$x-x^{2}+12x=36$$

13x ni olish uchun x va 12x ni birlashtirish.

$$13x-x^{2}=36$$

Ikkala tarafdan 36 ni ayirish.

$$13x-x^{2}-36=0$$

Polinomni standart shaklga keltirish uchun uni qayta tartiblang. Shartlarni eng yuqoridan eng pastki qiymat ko'rsatgichiga joylashtirish.

$$-x^{2}+13x-36=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon -x^{2}+ax+bx-36 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=13 ab=-\left(-36\right)=36$$

ab musbat boʻlganda, a va b da bir xil belgi bor. a+b musbat boʻlganda, a va b ikkisi ham musbat. 36-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,36 2,18 3,12 4,9 6,6$$

Har bir juftlik yigʻindisini hisoblang.

$$1+36=37 2+18=20 3+12=15 4+9=13 6+6=12$$

Yechim – 13 yigʻindisini beruvchi juftlik.

$$a=9 b=4$$

-x^{2}+13x-36 ni \left(-x^{2}+9x\right)+\left(4x-36\right) sifatida qaytadan yozish.

$$\left(-x^{2}+9x\right)+\left(4x-36\right)$$

Birinchi guruhda -x ni va ikkinchi guruhda 4 ni faktordan chiqaring.

$$-x\left(x-9\right)+4\left(x-9\right)$$

Distributiv funktsiyasidan foydalangan holda x-9 umumiy terminini chiqaring.

$$\left(x-9\right)\left(-x+4\right)$$

Tenglamani yechish uchun x-9=0 va -x+4=0 ni yeching.

$$x=9 x=4$$

Kvadrat tenglamalaridan foydalanish qoidalari

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(x-6\right)^{2} kengaytirilishi uchun ishlating.

$$x=x^{2}-12x+36$$

Ikkala tarafdan x^{2} ni ayirish.

$$x-x^{2}=-12x+36$$

12x ni ikki tarafga qo’shing.

$$x-x^{2}+12x=36$$

13x ni olish uchun x va 12x ni birlashtirish.

$$13x-x^{2}=36$$

Ikkala tarafdan 36 ni ayirish.

$$13x-x^{2}-36=0$$

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.

$$-x^{2}+13x-36=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} -1 ni a, 13 ni b va -36 ni c bilan almashtiring.

$$x=\frac{-13±\sqrt{13^{2}-4\left(-1\right)\left(-36\right)}}{2\left(-1\right)}$$

13 kvadratini chiqarish.

$$x=\frac{-13±\sqrt{169-4\left(-1\right)\left(-36\right)}}{2\left(-1\right)}$$

-4 ni -1 marotabaga ko'paytirish.

$$x=\frac{-13±\sqrt{169+4\left(-36\right)}}{2\left(-1\right)}$$

4 ni -36 marotabaga ko'paytirish.

$$x=\frac{-13±\sqrt{169-144}}{2\left(-1\right)}$$

169 ni -144 ga qo'shish.

$$x=\frac{-13±\sqrt{25}}{2\left(-1\right)}$$

25 ning kvadrat ildizini chiqarish.

$$x=\frac{-13±5}{2\left(-1\right)}$$

2 ni -1 marotabaga ko'paytirish.

$$x=\frac{-13±5}{-2}$$

x=\frac{-13±5}{-2} tenglamasini yeching, bunda ± musbat. -13 ni 5 ga qo'shish.

$$x=-\frac{8}{-2}$$

-8 ni -2 ga bo'lish.

$$x=4$$

x=\frac{-13±5}{-2} tenglamasini yeching, bunda ± manfiy. -13 dan 5 ni ayirish.

$$x=-\frac{18}{-2}$$

-18 ni -2 ga bo'lish.

$$x=9$$

Tenglama yechildi.

$$x=4 x=9$$

Kvadratni yakunlash bo'yicha qadamlar

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(x-6\right)^{2} kengaytirilishi uchun ishlating.

$$x=x^{2}-12x+36$$

Ikkala tarafdan x^{2} ni ayirish.

$$x-x^{2}=-12x+36$$

12x ni ikki tarafga qo’shing.

$$x-x^{2}+12x=36$$

13x ni olish uchun x va 12x ni birlashtirish.

$$13x-x^{2}=36$$

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

$$-x^{2}+13x=36$$

Ikki tarafini -1 ga bo‘ling.

$$\frac{-x^{2}+13x}{-1}=\frac{36}{-1}$$

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

$$x^{2}+\frac{13}{-1}x=\frac{36}{-1}$$

13 ni -1 ga bo'lish.

$$x^{2}-13x=\frac{36}{-1}$$

36 ni -1 ga bo'lish.

$$x^{2}-13x=-36$$

-13 ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{13}{2} olish uchun. Keyin, -\frac{13}{2} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-13x+\left(-\frac{13}{2}\right)^{2}=-36+\left(-\frac{13}{2}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{13}{2} kvadratini chiqarish.

$$x^{2}-13x+\frac{169}{4}=-36+\frac{169}{4}$$

-36 ni \frac{169}{4} ga qo'shish.

$$x^{2}-13x+\frac{169}{4}=\frac{25}{4}$$

x^{2}-13x+\frac{169}{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.

$$\left(x-\frac{13}{2}\right)^{2}=\frac{25}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{13}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Qisqartirish.

$$x-\frac{13}{2}=\frac{5}{2} x-\frac{13}{2}=-\frac{5}{2}$$

\frac{13}{2} ni tenglamaning ikkala tarafiga qo'shish.

$$x=9 x=4$$