(2x+1)^2=3-x eching

$$( 2 x + 1 ) ^ { 2 } = 3 - x$$

x uchun yechish

$$x=\frac{\sqrt{57}-5}{8}\approx 0,318729304$$

$$x=\frac{-\sqrt{57}-5}{8}\approx -1,568729304$$

Kvadrat tenglamalaridan foydalanish qoidalari

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(2x+1\right)^{2} kengaytirilishi uchun ishlating.

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

Ikkala tarafdan 3 ni ayirish.

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

-2 olish uchun 1 dan 3 ni ayirish.

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

x ni ikki tarafga qo’shing.

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

5x ni olish uchun 4x va x ni birlashtirish.

$$4x^{2}+5x-2=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 4 ni a, 5 ni b va -2 ni c bilan almashtiring.

$$x=\frac{-5±\sqrt{5^{2}-4\times 4\left(-2\right)}}{2\times 4}$$

5 kvadratini chiqarish.

$$x=\frac{-5±\sqrt{25-4\times 4\left(-2\right)}}{2\times 4}$$

-4 ni 4 marotabaga ko'paytirish.

$$x=\frac{-5±\sqrt{25-16\left(-2\right)}}{2\times 4}$$

-16 ni -2 marotabaga ko'paytirish.

$$x=\frac{-5±\sqrt{25+32}}{2\times 4}$$

25 ni 32 ga qo'shish.

$$x=\frac{-5±\sqrt{57}}{2\times 4}$$

2 ni 4 marotabaga ko'paytirish.

$$x=\frac{-5±\sqrt{57}}{8}$$

x=\frac{-5±\sqrt{57}}{8} tenglamasini yeching, bunda ± musbat. -5 ni \sqrt{57} ga qo'shish.

$$x=\frac{\sqrt{57}-5}{8}$$

x=\frac{-5±\sqrt{57}}{8} tenglamasini yeching, bunda ± manfiy. -5 dan \sqrt{57} ni ayirish.

$$x=\frac{-\sqrt{57}-5}{8}$$

Tenglama yechildi.

$$x=\frac{\sqrt{57}-5}{8} x=\frac{-\sqrt{57}-5}{8}$$

Kvadratni yakunlash bo'yicha qadamlar

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(2x+1\right)^{2} kengaytirilishi uchun ishlating.

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

x ni ikki tarafga qo’shing.

$$4x^{2}+4x+1+x=3$$

5x ni olish uchun 4x va x ni birlashtirish.

$$4x^{2}+5x+1=3$$

Ikkala tarafdan 1 ni ayirish.

$$4x^{2}+5x=3-1$$

2 olish uchun 3 dan 1 ni ayirish.

$$4x^{2}+5x=2$$

Ikki tarafini 4 ga bo‘ling.

$$\frac{4x^{2}+5x}{4}=\frac{2}{4}$$

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

$$x^{2}+\frac{5}{4}x=\frac{2}{4}$$

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

$$x^{2}+\frac{5}{4}x=\frac{1}{2}$$

\frac{5}{4} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{5}{8} olish uchun. Keyin, \frac{5}{8} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{5}{4}x+\left(\frac{5}{8}\right)^{2}=\frac{1}{2}+\left(\frac{5}{8}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{5}{8} kvadratini chiqarish.

$$x^{2}+\frac{5}{4}x+\frac{25}{64}=\frac{1}{2}+\frac{25}{64}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{1}{2} ni \frac{25}{64} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{5}{4}x+\frac{25}{64}=\frac{57}{64}$$

x^{2}+\frac{5}{4}x+\frac{25}{64} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{5}{8}\right)^{2}=\frac{57}{64}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{5}{8}\right)^{2}}=\sqrt{\frac{57}{64}}$$

Qisqartirish.

$$x+\frac{5}{8}=\frac{\sqrt{57}}{8} x+\frac{5}{8}=-\frac{\sqrt{57}}{8}$$

Tenglamaning ikkala tarafidan \frac{5}{8} ni ayirish.

$$x=\frac{\sqrt{57}-5}{8} x=\frac{-\sqrt{57}-5}{8}$$