10^2+x^2=8^2-(12-x)^2 eching

$$10 ^ { 2 } + x ^ { 2 } = 8 ^ { 2 } - ( 12 - x ) ^ { 2 }$$

x uchun yechish (complex solution)

$$x=6+3\sqrt{6}i\approx 6+7,348469228i$$

$$x=-3\sqrt{6}i+6\approx 6-7,348469228i$$

Kvadrat tenglamalaridan foydalanish qoidalari

2 daraja ko‘rsatkichini 10 ga hisoblang va 100 ni qiymatni oling.

$$100+x^{2}=8^{2}-\left(12-x\right)^{2}$$

2 daraja ko‘rsatkichini 8 ga hisoblang va 64 ni qiymatni oling.

$$100+x^{2}=64-\left(12-x\right)^{2}$$

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(12-x\right)^{2} kengaytirilishi uchun ishlating.

$$100+x^{2}=64-\left(144-24x+x^{2}\right)$$

144-24x+x^{2} teskarisini topish uchun har birining teskarisini toping.

$$100+x^{2}=64-144+24x-x^{2}$$

-80 olish uchun 64 dan 144 ni ayirish.

$$100+x^{2}=-80+24x-x^{2}$$

Ikkala tarafdan -80 ni ayirish.

$$100+x^{2}-\left(-80\right)=24x-x^{2}$$

-80 ning teskarisi 80 ga teng.

$$100+x^{2}+80=24x-x^{2}$$

Ikkala tarafdan 24x ni ayirish.

$$100+x^{2}+80-24x=-x^{2}$$

180 olish uchun 100 va 80'ni qo'shing.

$$180+x^{2}-24x=-x^{2}$$

x^{2} ni ikki tarafga qo’shing.

$$180+x^{2}-24x+x^{2}=0$$

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

$$180+2x^{2}-24x=0$$

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.

$$2x^{2}-24x+180=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 2 ni a, -24 ni b va 180 ni c bilan almashtiring.

$$x=\frac{-\left(-24\right)±\sqrt{\left(-24\right)^{2}-4\times 2\times 180}}{2\times 2}$$

-24 kvadratini chiqarish.

$$x=\frac{-\left(-24\right)±\sqrt{576-4\times 2\times 180}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-\left(-24\right)±\sqrt{576-8\times 180}}{2\times 2}$$

-8 ni 180 marotabaga ko'paytirish.

$$x=\frac{-\left(-24\right)±\sqrt{576-1440}}{2\times 2}$$

576 ni -1440 ga qo'shish.

$$x=\frac{-\left(-24\right)±\sqrt{-864}}{2\times 2}$$

-864 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-24\right)±12\sqrt{6}i}{2\times 2}$$

-24 ning teskarisi 24 ga teng.

$$x=\frac{24±12\sqrt{6}i}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{24±12\sqrt{6}i}{4}$$

x=\frac{24±12\sqrt{6}i}{4} tenglamasini yeching, bunda ± musbat. 24 ni 12i\sqrt{6} ga qo'shish.

$$x=\frac{24+12\sqrt{6}i}{4}$$

24+12i\sqrt{6} ni 4 ga bo'lish.

$$x=6+3\sqrt{6}i$$

x=\frac{24±12\sqrt{6}i}{4} tenglamasini yeching, bunda ± manfiy. 24 dan 12i\sqrt{6} ni ayirish.

$$x=\frac{-12\sqrt{6}i+24}{4}$$

24-12i\sqrt{6} ni 4 ga bo'lish.

$$x=-3\sqrt{6}i+6$$

Tenglama yechildi.

$$x=6+3\sqrt{6}i x=-3\sqrt{6}i+6$$

Kvadratni yakunlash bo'yicha qadamlar

2 daraja ko‘rsatkichini 10 ga hisoblang va 100 ni qiymatni oling.

$$100+x^{2}=8^{2}-\left(12-x\right)^{2}$$

2 daraja ko‘rsatkichini 8 ga hisoblang va 64 ni qiymatni oling.

$$100+x^{2}=64-\left(12-x\right)^{2}$$

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(12-x\right)^{2} kengaytirilishi uchun ishlating.

$$100+x^{2}=64-\left(144-24x+x^{2}\right)$$

144-24x+x^{2} teskarisini topish uchun har birining teskarisini toping.

$$100+x^{2}=64-144+24x-x^{2}$$

-80 olish uchun 64 dan 144 ni ayirish.

$$100+x^{2}=-80+24x-x^{2}$$

Ikkala tarafdan 24x ni ayirish.

$$100+x^{2}-24x=-80-x^{2}$$

x^{2} ni ikki tarafga qo’shing.

$$100+x^{2}-24x+x^{2}=-80$$

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

$$100+2x^{2}-24x=-80$$

Ikkala tarafdan 100 ni ayirish.

$$2x^{2}-24x=-80-100$$

-180 olish uchun -80 dan 100 ni ayirish.

$$2x^{2}-24x=-180$$

Ikki tarafini 2 ga bo‘ling.

$$\frac{2x^{2}-24x}{2}=-\frac{180}{2}$$

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

$$x^{2}+\left(-\frac{24}{2}\right)x=-\frac{180}{2}$$

-24 ni 2 ga bo'lish.

$$x^{2}-12x=-\frac{180}{2}$$

-180 ni 2 ga bo'lish.

$$x^{2}-12x=-90$$

-12 ni bo‘lish, x shartining koeffitsienti, 2 ga -6 olish uchun. Keyin, -6 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-12x+\left(-6\right)^{2}=-90+\left(-6\right)^{2}$$

-6 kvadratini chiqarish.

$$x^{2}-12x+36=-90+36$$

-90 ni 36 ga qo'shish.

$$x^{2}-12x+36=-54$$

x^{2}-12x+36 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-6\right)^{2}=-54$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-6\right)^{2}}=\sqrt{-54}$$

Qisqartirish.

$$x-6=3\sqrt{6}i x-6=-3\sqrt{6}i$$

6 ni tenglamaning ikkala tarafiga qo'shish.

$$x=6+3\sqrt{6}i x=-3\sqrt{6}i+6$$