6^2=x(x-3) eching

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

x uchun yechish

$$x = \frac{3 \sqrt{17} + 3}{2} \approx 7,684658438$$

$$x=\frac{3-3\sqrt{17}}{2}\approx -4,684658438$$

Kvadrat tenglamalaridan foydalanish qoidalari

2 daraja ko‘rsatkichini 6 ga hisoblang va 36 ni qiymatni oling.

$$36=x\left(x-3\right)$$

x ga x-3 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

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

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

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

Ikkala tarafdan 36 ni ayirish.

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

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 -36 ni c bilan almashtiring.

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

-3 kvadratini chiqarish.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-36\right)}}{2}$$

-4 ni -36 marotabaga ko'paytirish.

$$x=\frac{-\left(-3\right)±\sqrt{9+144}}{2}$$

9 ni 144 ga qo'shish.

$$x=\frac{-\left(-3\right)±\sqrt{153}}{2}$$

153 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-3\right)±3\sqrt{17}}{2}$$

-3 ning teskarisi 3 ga teng.

$$x=\frac{3±3\sqrt{17}}{2}$$

x=\frac{3±3\sqrt{17}}{2} tenglamasini yeching, bunda ± musbat. 3 ni 3\sqrt{17} ga qo'shish.

$$x=\frac{3\sqrt{17}+3}{2}$$

x=\frac{3±3\sqrt{17}}{2} tenglamasini yeching, bunda ± manfiy. 3 dan 3\sqrt{17} ni ayirish.

$$x=\frac{3-3\sqrt{17}}{2}$$

Tenglama yechildi.

$$x=\frac{3\sqrt{17}+3}{2} x=\frac{3-3\sqrt{17}}{2}$$

Kvadratni yakunlash bo'yicha qadamlar

2 daraja ko‘rsatkichini 6 ga hisoblang va 36 ni qiymatni oling.

$$36=x\left(x-3\right)$$

x ga x-3 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

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

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

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

-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.

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=36+\left(-\frac{3}{2}\right)^{2}$$

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

$$x^{2}-3x+\frac{9}{4}=36+\frac{9}{4}$$

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

$$x^{2}-3x+\frac{9}{4}=\frac{153}{4}$$

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.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{153}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{153}{4}}$$

Qisqartirish.

$$x-\frac{3}{2}=\frac{3\sqrt{17}}{2} x-\frac{3}{2}=-\frac{3\sqrt{17}}{2}$$

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

$$x=\frac{3\sqrt{17}+3}{2} x=\frac{3-3\sqrt{17}}{2}$$