(x+3)-10-(x-2)^2=(x-1)^2-(x-5)^2+4+x eching

$$( x + 3 ) - 10 - ( x - 2 ) ^ { 2 } = ( x - 1 ) ^ { 2 } - ( x - 5 ) ^ { 2 } + 4 + x$$

x uchun yechish

$$x=\sqrt{13}-2\approx 1,605551275$$

$$x=-\left(\sqrt{13}+2\right)\approx -5,605551275$$

Kvadrat tenglamalaridan foydalanish qoidalari

-7 olish uchun 3 dan 10 ni ayirish.

$$x-7-\left(x-2\right)^{2}=\left(x-1\right)^{2}-\left(x-5\right)^{2}+4+x$$

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

$$x-7-\left(x^{2}-4x+4\right)=\left(x-1\right)^{2}-\left(x-5\right)^{2}+4+x$$

x^{2}-4x+4 teskarisini topish uchun har birining teskarisini toping.

$$x-7-x^{2}+4x-4=\left(x-1\right)^{2}-\left(x-5\right)^{2}+4+x$$

5x ni olish uchun x va 4x ni birlashtirish.

$$5x-7-x^{2}-4=\left(x-1\right)^{2}-\left(x-5\right)^{2}+4+x$$

-11 olish uchun -7 dan 4 ni ayirish.

$$5x-11-x^{2}=\left(x-1\right)^{2}-\left(x-5\right)^{2}+4+x$$

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

$$5x-11-x^{2}=x^{2}-2x+1-\left(x-5\right)^{2}+4+x$$

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(x-5\right)^{2} kengaytirilishi uchun ishlating.

$$5x-11-x^{2}=x^{2}-2x+1-\left(x^{2}-10x+25\right)+4+x$$

x^{2}-10x+25 teskarisini topish uchun har birining teskarisini toping.

$$5x-11-x^{2}=x^{2}-2x+1-x^{2}+10x-25+4+x$$

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

$$5x-11-x^{2}=-2x+1+10x-25+4+x$$

8x ni olish uchun -2x va 10x ni birlashtirish.

$$5x-11-x^{2}=8x+1-25+4+x$$

-24 olish uchun 1 dan 25 ni ayirish.

$$5x-11-x^{2}=8x-24+4+x$$

-20 olish uchun -24 va 4'ni qo'shing.

$$5x-11-x^{2}=8x-20+x$$

9x ni olish uchun 8x va x ni birlashtirish.

$$5x-11-x^{2}=9x-20$$

Ikkala tarafdan 9x ni ayirish.

$$5x-11-x^{2}-9x=-20$$

-4x ni olish uchun 5x va -9x ni birlashtirish.

$$-4x-11-x^{2}=-20$$

20 ni ikki tarafga qo’shing.

$$-4x-11-x^{2}+20=0$$

9 olish uchun -11 va 20'ni qo'shing.

$$-4x+9-x^{2}=0$$

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.

$$-x^{2}-4x+9=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} -1 ni a, -4 ni b va 9 ni c bilan almashtiring.

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

-4 kvadratini chiqarish.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-1\right)\times 9}}{2\left(-1\right)}$$

-4 ni -1 marotabaga ko'paytirish.

$$x=\frac{-\left(-4\right)±\sqrt{16+4\times 9}}{2\left(-1\right)}$$

4 ni 9 marotabaga ko'paytirish.

$$x=\frac{-\left(-4\right)±\sqrt{16+36}}{2\left(-1\right)}$$

16 ni 36 ga qo'shish.

$$x=\frac{-\left(-4\right)±\sqrt{52}}{2\left(-1\right)}$$

52 ning kvadrat ildizini chiqarish.

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

-4 ning teskarisi 4 ga teng.

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

2 ni -1 marotabaga ko'paytirish.

$$x=\frac{4±2\sqrt{13}}{-2}$$

x=\frac{4±2\sqrt{13}}{-2} tenglamasini yeching, bunda ± musbat. 4 ni 2\sqrt{13} ga qo'shish.

$$x=\frac{2\sqrt{13}+4}{-2}$$

4+2\sqrt{13} ni -2 ga bo'lish.

$$x=-\left(\sqrt{13}+2\right)$$

x=\frac{4±2\sqrt{13}}{-2} tenglamasini yeching, bunda ± manfiy. 4 dan 2\sqrt{13} ni ayirish.

$$x=\frac{4-2\sqrt{13}}{-2}$$

4-2\sqrt{13} ni -2 ga bo'lish.

$$x=\sqrt{13}-2$$

Tenglama yechildi.

$$x=-\left(\sqrt{13}+2\right) x=\sqrt{13}-2$$

Kvadratni yakunlash bo'yicha qadamlar

-7 olish uchun 3 dan 10 ni ayirish.

$$x-7-\left(x-2\right)^{2}=\left(x-1\right)^{2}-\left(x-5\right)^{2}+4+x$$

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

$$x-7-\left(x^{2}-4x+4\right)=\left(x-1\right)^{2}-\left(x-5\right)^{2}+4+x$$

x^{2}-4x+4 teskarisini topish uchun har birining teskarisini toping.

$$x-7-x^{2}+4x-4=\left(x-1\right)^{2}-\left(x-5\right)^{2}+4+x$$

5x ni olish uchun x va 4x ni birlashtirish.

$$5x-7-x^{2}-4=\left(x-1\right)^{2}-\left(x-5\right)^{2}+4+x$$

-11 olish uchun -7 dan 4 ni ayirish.

$$5x-11-x^{2}=\left(x-1\right)^{2}-\left(x-5\right)^{2}+4+x$$

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

$$5x-11-x^{2}=x^{2}-2x+1-\left(x-5\right)^{2}+4+x$$

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(x-5\right)^{2} kengaytirilishi uchun ishlating.

$$5x-11-x^{2}=x^{2}-2x+1-\left(x^{2}-10x+25\right)+4+x$$

x^{2}-10x+25 teskarisini topish uchun har birining teskarisini toping.

$$5x-11-x^{2}=x^{2}-2x+1-x^{2}+10x-25+4+x$$

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

$$5x-11-x^{2}=-2x+1+10x-25+4+x$$

8x ni olish uchun -2x va 10x ni birlashtirish.

$$5x-11-x^{2}=8x+1-25+4+x$$

-24 olish uchun 1 dan 25 ni ayirish.

$$5x-11-x^{2}=8x-24+4+x$$

-20 olish uchun -24 va 4'ni qo'shing.

$$5x-11-x^{2}=8x-20+x$$

9x ni olish uchun 8x va x ni birlashtirish.

$$5x-11-x^{2}=9x-20$$

Ikkala tarafdan 9x ni ayirish.

$$5x-11-x^{2}-9x=-20$$

-4x ni olish uchun 5x va -9x ni birlashtirish.

$$-4x-11-x^{2}=-20$$

11 ni ikki tarafga qo’shing.

$$-4x-x^{2}=-20+11$$

-9 olish uchun -20 va 11'ni qo'shing.

$$-4x-x^{2}=-9$$

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

$$-x^{2}-4x=-9$$

Ikki tarafini -1 ga bo‘ling.

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

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

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

-4 ni -1 ga bo'lish.

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

-9 ni -1 ga bo'lish.

$$x^{2}+4x=9$$

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

$$x^{2}+4x+2^{2}=9+2^{2}$$

2 kvadratini chiqarish.

$$x^{2}+4x+4=9+4$$

9 ni 4 ga qo'shish.

$$x^{2}+4x+4=13$$

x^{2}+4x+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.

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

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{13}$$

Qisqartirish.

$$x+2=\sqrt{13} x+2=-\sqrt{13}$$

Tenglamaning ikkala tarafidan 2 ni ayirish.

$$x=\sqrt{13}-2 x=-\sqrt{13}-2$$