(x-6)^2=2x^2-15x+38 eching

$$( x - 6 ) ^ { 2 } = 2 x ^ { 2 } - 15 x + 38$$

x uchun yechish

$$x=1$$

$$x=2$$

Guruhlash orqali faktorlash qadamlari

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

$$x^{2}-12x+36=2x^{2}-15x+38$$

Ikkala tarafdan 2x^{2} ni ayirish.

$$x^{2}-12x+36-2x^{2}=-15x+38$$

-x^{2} ni olish uchun x^{2} va -2x^{2} ni birlashtirish.

$$-x^{2}-12x+36=-15x+38$$

15x ni ikki tarafga qo’shing.

$$-x^{2}-12x+36+15x=38$$

3x ni olish uchun -12x va 15x ni birlashtirish.

$$-x^{2}+3x+36=38$$

Ikkala tarafdan 38 ni ayirish.

$$-x^{2}+3x+36-38=0$$

-2 olish uchun 36 dan 38 ni ayirish.

$$-x^{2}+3x-2=0$$

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

$$a+b=3 ab=-\left(-2\right)=2$$

ab musbat boʻlganda, a va b da bir xil belgi bor. a+b musbat boʻlganda, a va b ikkisi ham musbat. Faqat bundan juftlik tizim yechimidir.

$$a=2 b=1$$

-x^{2}+3x-2 ni \left(-x^{2}+2x\right)+\left(x-2\right) sifatida qaytadan yozish.

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

-x^{2}+2x ichida -x ni ajrating.

$$-x\left(x-2\right)+x-2$$

Distributiv funktsiyasidan foydalangan holda x-2 umumiy terminini chiqaring.

$$\left(x-2\right)\left(-x+1\right)$$

Tenglamani yechish uchun x-2=0 va -x+1=0 ni yeching.

$$x=2 x=1$$

Kvadrat tenglamalaridan foydalanish qoidalari

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

$$x^{2}-12x+36=2x^{2}-15x+38$$

Ikkala tarafdan 2x^{2} ni ayirish.

$$x^{2}-12x+36-2x^{2}=-15x+38$$

-x^{2} ni olish uchun x^{2} va -2x^{2} ni birlashtirish.

$$-x^{2}-12x+36=-15x+38$$

15x ni ikki tarafga qo’shing.

$$-x^{2}-12x+36+15x=38$$

3x ni olish uchun -12x va 15x ni birlashtirish.

$$-x^{2}+3x+36=38$$

Ikkala tarafdan 38 ni ayirish.

$$-x^{2}+3x+36-38=0$$

-2 olish uchun 36 dan 38 ni ayirish.

$$-x^{2}+3x-2=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} -1 ni a, 3 ni b va -2 ni c bilan almashtiring.

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

3 kvadratini chiqarish.

$$x=\frac{-3±\sqrt{9-4\left(-1\right)\left(-2\right)}}{2\left(-1\right)}$$

-4 ni -1 marotabaga ko'paytirish.

$$x=\frac{-3±\sqrt{9+4\left(-2\right)}}{2\left(-1\right)}$$

4 ni -2 marotabaga ko'paytirish.

$$x=\frac{-3±\sqrt{9-8}}{2\left(-1\right)}$$

9 ni -8 ga qo'shish.

$$x=\frac{-3±\sqrt{1}}{2\left(-1\right)}$$

1 ning kvadrat ildizini chiqarish.

$$x=\frac{-3±1}{2\left(-1\right)}$$

2 ni -1 marotabaga ko'paytirish.

$$x=\frac{-3±1}{-2}$$

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

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

-2 ni -2 ga bo'lish.

$$x=1$$

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

$$x=-\frac{4}{-2}$$

-4 ni -2 ga bo'lish.

$$x=2$$

Tenglama yechildi.

$$x=1 x=2$$

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.

$$x^{2}-12x+36=2x^{2}-15x+38$$

Ikkala tarafdan 2x^{2} ni ayirish.

$$x^{2}-12x+36-2x^{2}=-15x+38$$

-x^{2} ni olish uchun x^{2} va -2x^{2} ni birlashtirish.

$$-x^{2}-12x+36=-15x+38$$

15x ni ikki tarafga qo’shing.

$$-x^{2}-12x+36+15x=38$$

3x ni olish uchun -12x va 15x ni birlashtirish.

$$-x^{2}+3x+36=38$$

Ikkala tarafdan 36 ni ayirish.

$$-x^{2}+3x=38-36$$

2 olish uchun 38 dan 36 ni ayirish.

$$-x^{2}+3x=2$$

Ikki tarafini -1 ga bo‘ling.

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

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

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

3 ni -1 ga bo'lish.

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

2 ni -1 ga bo'lish.

$$x^{2}-3x=-2$$

-3 ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{3}{2} olish uchun. Keyin, -\frac{3}{2} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{3}{2} kvadratini chiqarish.

$$x^{2}-3x+\frac{9}{4}=-2+\frac{9}{4}$$

-2 ni \frac{9}{4} ga qo'shish.

$$x^{2}-3x+\frac{9}{4}=\frac{1}{4}$$

x^{2}-3x+\frac{9}{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.

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

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

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

Qisqartirish.

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

\frac{3}{2} ni tenglamaning ikkala tarafiga qo'shish.

$$x=2 x=1$$