3^2=(sqrt{3})^2+(3-x)^2 eching

$$3 ^ { 2 } = ( \sqrt { 3 } ) ^ { 2 } + ( 3 - x ) ^ { 2 }$$

x uchun yechish

$$x=3-\sqrt{6}\approx 0,550510257$$

$$x=\sqrt{6}+3\approx 5,449489743$$

Kvadrat tenglamalaridan foydalanish qoidalari

2 daraja ko‘rsatkichini 3 ga hisoblang va 9 ni qiymatni oling.

$$9=\left(\sqrt{3}\right)^{2}+\left(3-x\right)^{2}$$

\sqrt{3} kvadrati – 3.

$$9=3+\left(3-x\right)^{2}$$

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(3-x\right)^{2} kengaytirilishi uchun ishlating.

$$9=3+9-6x+x^{2}$$

12 olish uchun 3 va 9'ni qo'shing.

$$9=12-6x+x^{2}$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$12-6x+x^{2}=9$$

Ikkala tarafdan 9 ni ayirish.

$$12-6x+x^{2}-9=0$$

3 olish uchun 12 dan 9 ni ayirish.

$$3-6x+x^{2}=0$$

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.

$$x^{2}-6x+3=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, -6 ni b va 3 ni c bilan almashtiring.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 3}}{2}$$

-6 kvadratini chiqarish.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 3}}{2}$$

-4 ni 3 marotabaga ko'paytirish.

$$x=\frac{-\left(-6\right)±\sqrt{36-12}}{2}$$

36 ni -12 ga qo'shish.

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

24 ning kvadrat ildizini chiqarish.

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

-6 ning teskarisi 6 ga teng.

$$x=\frac{6±2\sqrt{6}}{2}$$

x=\frac{6±2\sqrt{6}}{2} tenglamasini yeching, bunda ± musbat. 6 ni 2\sqrt{6} ga qo'shish.

$$x=\frac{2\sqrt{6}+6}{2}$$

6+2\sqrt{6} ni 2 ga bo'lish.

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

x=\frac{6±2\sqrt{6}}{2} tenglamasini yeching, bunda ± manfiy. 6 dan 2\sqrt{6} ni ayirish.

$$x=\frac{6-2\sqrt{6}}{2}$$

6-2\sqrt{6} ni 2 ga bo'lish.

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

Tenglama yechildi.

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

Kvadratni yakunlash bo'yicha qadamlar

2 daraja ko‘rsatkichini 3 ga hisoblang va 9 ni qiymatni oling.

$$9=\left(\sqrt{3}\right)^{2}+\left(3-x\right)^{2}$$

\sqrt{3} kvadrati – 3.

$$9=3+\left(3-x\right)^{2}$$

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(3-x\right)^{2} kengaytirilishi uchun ishlating.

$$9=3+9-6x+x^{2}$$

12 olish uchun 3 va 9'ni qo'shing.

$$9=12-6x+x^{2}$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$12-6x+x^{2}=9$$

Ikkala tarafdan 12 ni ayirish.

$$-6x+x^{2}=9-12$$

-3 olish uchun 9 dan 12 ni ayirish.

$$-6x+x^{2}=-3$$

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

$$x^{2}-6x=-3$$

-6 ni bo‘lish, x shartining koeffitsienti, 2 ga -3 olish uchun. Keyin, -3 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-6x+\left(-3\right)^{2}=-3+\left(-3\right)^{2}$$

-3 kvadratini chiqarish.

$$x^{2}-6x+9=-3+9$$

-3 ni 9 ga qo'shish.

$$x^{2}-6x+9=6$$

x^{2}-6x+9 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-3\right)^{2}=6$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{6}$$

Qisqartirish.

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

3 ni tenglamaning ikkala tarafiga qo'shish.

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