left(2xright)^2-2x-3=0 eching

$${ \left(2x \right) }^{ 2 } -2x-3=0$$

x uchun yechish

$$x = \frac{\sqrt{13} + 1}{4} \approx 1,151387819$$

$$x=\frac{1-\sqrt{13}}{4}\approx -0,651387819$$

Kvadrat tenglamalaridan foydalanish qoidalari

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

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

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

$$4x^{2}-2x-3=0$$

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

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

-2 kvadratini chiqarish.

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

-4 ni 4 marotabaga ko'paytirish.

$$x=\frac{-\left(-2\right)±\sqrt{4-16\left(-3\right)}}{2\times 4}$$

-16 ni -3 marotabaga ko'paytirish.

$$x=\frac{-\left(-2\right)±\sqrt{4+48}}{2\times 4}$$

4 ni 48 ga qo'shish.

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

52 ning kvadrat ildizini chiqarish.

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

-2 ning teskarisi 2 ga teng.

$$x=\frac{2±2\sqrt{13}}{2\times 4}$$

2 ni 4 marotabaga ko'paytirish.

$$x=\frac{2±2\sqrt{13}}{8}$$

x=\frac{2±2\sqrt{13}}{8} tenglamasini yeching, bunda ± musbat. 2 ni 2\sqrt{13} ga qo'shish.

$$x=\frac{2\sqrt{13}+2}{8}$$

2+2\sqrt{13} ni 8 ga bo'lish.

$$x=\frac{\sqrt{13}+1}{4}$$

x=\frac{2±2\sqrt{13}}{8} tenglamasini yeching, bunda ± manfiy. 2 dan 2\sqrt{13} ni ayirish.

$$x=\frac{2-2\sqrt{13}}{8}$$

2-2\sqrt{13} ni 8 ga bo'lish.

$$x=\frac{1-\sqrt{13}}{4}$$

Tenglama yechildi.

$$x=\frac{\sqrt{13}+1}{4} x=\frac{1-\sqrt{13}}{4}$$

Kvadratni yakunlash bo'yicha qadamlar

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

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

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

$$4x^{2}-2x-3=0$$

3 ni ikki tarafga qo’shing. Har qanday songa nolni qo‘shsangiz, o‘zi chiqadi.

$$4x^{2}-2x=3$$

Ikki tarafini 4 ga bo‘ling.

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

4 ga bo'lish 4 ga ko'paytirishni bekor qiladi.

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

\frac{-2}{4} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}-\frac{1}{2}x=\frac{3}{4}$$

-\frac{1}{2} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{1}{4} olish uchun. Keyin, -\frac{1}{4} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{1}{4} kvadratini chiqarish.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{3}{4}+\frac{1}{16}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{3}{4} ni \frac{1}{16} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{13}{16}$$

x^{2}-\frac{1}{2}x+\frac{1}{16} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-\frac{1}{4}\right)^{2}=\frac{13}{16}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{\frac{13}{16}}$$

Qisqartirish.

$$x-\frac{1}{4}=\frac{\sqrt{13}}{4} x-\frac{1}{4}=-\frac{\sqrt{13}}{4}$$

\frac{1}{4} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{\sqrt{13}+1}{4} x=\frac{1-\sqrt{13}}{4}$$