{left(2sqrt{3}right)}^2=1/2xleft(6+xright) eching

$${ \left(2 \sqrt{ 3 } \right) }^{ 2 } = \frac{ 1 }{ 2 } x \left( 6+x \right)$$

x uchun yechish (complex solution)

$$x=\sqrt{33}-3\approx 2,744562647$$

$$x=-\left(\sqrt{33}+3\right)\approx -8,744562647$$

Kvadrat tenglamalaridan foydalanish qoidalari

\left(2\sqrt{3}\right)^{2} ni kengaytirish.

$$2^{2}\left(\sqrt{3}\right)^{2}=\frac{1}{2}x\left(6+x\right)$$

2 daraja ko‘rsatkichini 2 ga hisoblang va 4 ni qiymatni oling.

$$4\left(\sqrt{3}\right)^{2}=\frac{1}{2}x\left(6+x\right)$$

\sqrt{3} kvadrati – 3.

$$4\times 3=\frac{1}{2}x\left(6+x\right)$$

12 hosil qilish uchun 4 va 3 ni ko'paytirish.

$$12=\frac{1}{2}x\left(6+x\right)$$

\frac{1}{2}x ga 6+x ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$12=3x+\frac{1}{2}x^{2}$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$3x+\frac{1}{2}x^{2}=12$$

Ikkala tarafdan 12 ni ayirish.

$$3x+\frac{1}{2}x^{2}-12=0$$

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.

$$\frac{1}{2}x^{2}+3x-12=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} \frac{1}{2} ni a, 3 ni b va -12 ni c bilan almashtiring.

$$x=\frac{-3±\sqrt{3^{2}-4\times \frac{1}{2}\left(-12\right)}}{2\times \frac{1}{2}}$$

3 kvadratini chiqarish.

$$x=\frac{-3±\sqrt{9-4\times \frac{1}{2}\left(-12\right)}}{2\times \frac{1}{2}}$$

-4 ni \frac{1}{2} marotabaga ko'paytirish.

$$x=\frac{-3±\sqrt{9-2\left(-12\right)}}{2\times \frac{1}{2}}$$

-2 ni -12 marotabaga ko'paytirish.

$$x=\frac{-3±\sqrt{9+24}}{2\times \frac{1}{2}}$$

9 ni 24 ga qo'shish.

$$x=\frac{-3±\sqrt{33}}{2\times \frac{1}{2}}$$

2 ni \frac{1}{2} marotabaga ko'paytirish.

$$x=\frac{-3±\sqrt{33}}{1}$$

x=\frac{-3±\sqrt{33}}{1} tenglamasini yeching, bunda ± musbat. -3 ni \sqrt{33} ga qo'shish.

$$x=\frac{\sqrt{33}-3}{1}$$

-3+\sqrt{33} ni 1 ga bo'lish.

$$x=\sqrt{33}-3$$

x=\frac{-3±\sqrt{33}}{1} tenglamasini yeching, bunda ± manfiy. -3 dan \sqrt{33} ni ayirish.

$$x=\frac{-\sqrt{33}-3}{1}$$

-3-\sqrt{33} ni 1 ga bo'lish.

$$x=-\sqrt{33}-3$$

Tenglama yechildi.

$$x=\sqrt{33}-3 x=-\sqrt{33}-3$$

Kvadratni yakunlash bo'yicha qadamlar

\left(2\sqrt{3}\right)^{2} ni kengaytirish.

$$2^{2}\left(\sqrt{3}\right)^{2}=\frac{1}{2}x\left(6+x\right)$$

2 daraja ko‘rsatkichini 2 ga hisoblang va 4 ni qiymatni oling.

$$4\left(\sqrt{3}\right)^{2}=\frac{1}{2}x\left(6+x\right)$$

\sqrt{3} kvadrati – 3.

$$4\times 3=\frac{1}{2}x\left(6+x\right)$$

12 hosil qilish uchun 4 va 3 ni ko'paytirish.

$$12=\frac{1}{2}x\left(6+x\right)$$

\frac{1}{2}x ga 6+x ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$12=3x+\frac{1}{2}x^{2}$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$3x+\frac{1}{2}x^{2}=12$$

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

$$\frac{1}{2}x^{2}+3x=12$$

Ikkala tarafini 2 ga ko‘paytiring.

$$\frac{\frac{1}{2}x^{2}+3x}{\frac{1}{2}}=\frac{12}{\frac{1}{2}}$$

\frac{1}{2} ga bo'lish \frac{1}{2} ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{3}{\frac{1}{2}}x=\frac{12}{\frac{1}{2}}$$

3 ni \frac{1}{2} ga bo'lish 3 ga k'paytirish \frac{1}{2} ga qaytarish.

$$x^{2}+6x=\frac{12}{\frac{1}{2}}$$

12 ni \frac{1}{2} ga bo'lish 12 ga k'paytirish \frac{1}{2} ga qaytarish.

$$x^{2}+6x=24$$

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.

$$x^{2}+6x+3^{2}=24+3^{2}$$

3 kvadratini chiqarish.

$$x^{2}+6x+9=24+9$$

24 ni 9 ga qo'shish.

$$x^{2}+6x+9=33$$

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.

$$\left(x+3\right)^{2}=33$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+3\right)^{2}}=\sqrt{33}$$

Qisqartirish.

$$x+3=\sqrt{33} x+3=-\sqrt{33}$$

Tenglamaning ikkala tarafidan 3 ni ayirish.

$$x=\sqrt{33}-3 x=-\sqrt{33}-3$$

x uchun yechish

$$x=\sqrt{33}-3\approx 2,744562647$$

$$x=-\sqrt{33}-3\approx -8,744562647$$

Kvadrat tenglamalaridan foydalanish qoidalari

\left(2\sqrt{3}\right)^{2} ni kengaytirish.

$$2^{2}\left(\sqrt{3}\right)^{2}=\frac{1}{2}x\left(6+x\right)$$

2 daraja ko‘rsatkichini 2 ga hisoblang va 4 ni qiymatni oling.

$$4\left(\sqrt{3}\right)^{2}=\frac{1}{2}x\left(6+x\right)$$

\sqrt{3} kvadrati – 3.

$$4\times 3=\frac{1}{2}x\left(6+x\right)$$

12 hosil qilish uchun 4 va 3 ni ko'paytirish.

$$12=\frac{1}{2}x\left(6+x\right)$$

\frac{1}{2}x ga 6+x ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$12=3x+\frac{1}{2}x^{2}$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$3x+\frac{1}{2}x^{2}=12$$

Ikkala tarafdan 12 ni ayirish.

$$3x+\frac{1}{2}x^{2}-12=0$$

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.

$$\frac{1}{2}x^{2}+3x-12=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} \frac{1}{2} ni a, 3 ni b va -12 ni c bilan almashtiring.

$$x=\frac{-3±\sqrt{3^{2}-4\times \frac{1}{2}\left(-12\right)}}{2\times \frac{1}{2}}$$

3 kvadratini chiqarish.

$$x=\frac{-3±\sqrt{9-4\times \frac{1}{2}\left(-12\right)}}{2\times \frac{1}{2}}$$

-4 ni \frac{1}{2} marotabaga ko'paytirish.

$$x=\frac{-3±\sqrt{9-2\left(-12\right)}}{2\times \frac{1}{2}}$$

-2 ni -12 marotabaga ko'paytirish.

$$x=\frac{-3±\sqrt{9+24}}{2\times \frac{1}{2}}$$

9 ni 24 ga qo'shish.

$$x=\frac{-3±\sqrt{33}}{2\times \frac{1}{2}}$$

2 ni \frac{1}{2} marotabaga ko'paytirish.

$$x=\frac{-3±\sqrt{33}}{1}$$

x=\frac{-3±\sqrt{33}}{1} tenglamasini yeching, bunda ± musbat. -3 ni \sqrt{33} ga qo'shish.

$$x=\frac{\sqrt{33}-3}{1}$$

-3+\sqrt{33} ni 1 ga bo'lish.

$$x=\sqrt{33}-3$$

x=\frac{-3±\sqrt{33}}{1} tenglamasini yeching, bunda ± manfiy. -3 dan \sqrt{33} ni ayirish.

$$x=\frac{-\sqrt{33}-3}{1}$$

-3-\sqrt{33} ni 1 ga bo'lish.

$$x=-\sqrt{33}-3$$

Tenglama yechildi.

$$x=\sqrt{33}-3 x=-\sqrt{33}-3$$

Kvadratni yakunlash bo'yicha qadamlar

\left(2\sqrt{3}\right)^{2} ni kengaytirish.

$$2^{2}\left(\sqrt{3}\right)^{2}=\frac{1}{2}x\left(6+x\right)$$

2 daraja ko‘rsatkichini 2 ga hisoblang va 4 ni qiymatni oling.

$$4\left(\sqrt{3}\right)^{2}=\frac{1}{2}x\left(6+x\right)$$

\sqrt{3} kvadrati – 3.

$$4\times 3=\frac{1}{2}x\left(6+x\right)$$

12 hosil qilish uchun 4 va 3 ni ko'paytirish.

$$12=\frac{1}{2}x\left(6+x\right)$$

\frac{1}{2}x ga 6+x ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$12=3x+\frac{1}{2}x^{2}$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$3x+\frac{1}{2}x^{2}=12$$

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

$$\frac{1}{2}x^{2}+3x=12$$

Ikkala tarafini 2 ga ko‘paytiring.

$$\frac{\frac{1}{2}x^{2}+3x}{\frac{1}{2}}=\frac{12}{\frac{1}{2}}$$

\frac{1}{2} ga bo'lish \frac{1}{2} ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{3}{\frac{1}{2}}x=\frac{12}{\frac{1}{2}}$$

3 ni \frac{1}{2} ga bo'lish 3 ga k'paytirish \frac{1}{2} ga qaytarish.

$$x^{2}+6x=\frac{12}{\frac{1}{2}}$$

12 ni \frac{1}{2} ga bo'lish 12 ga k'paytirish \frac{1}{2} ga qaytarish.

$$x^{2}+6x=24$$

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.

$$x^{2}+6x+3^{2}=24+3^{2}$$

3 kvadratini chiqarish.

$$x^{2}+6x+9=24+9$$

24 ni 9 ga qo'shish.

$$x^{2}+6x+9=33$$

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.

$$\left(x+3\right)^{2}=33$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+3\right)^{2}}=\sqrt{33}$$

Qisqartirish.

$$x+3=\sqrt{33} x+3=-\sqrt{33}$$

Tenglamaning ikkala tarafidan 3 ni ayirish.

$$x=\sqrt{33}-3 x=-\sqrt{33}-3$$