(2y+3)^2+y^2=4 eching

$$( 2 y + 3 ) ^ { 2 } + y ^ { 2 } = 4$$

y uchun yechish

$$y=\frac{\sqrt{11}-6}{5}\approx -0,536675042$$

$$y=\frac{-\sqrt{11}-6}{5}\approx -1,863324958$$

Kvadrat tenglamalaridan foydalanish qoidalari

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

$$4y^{2}+12y+9+y^{2}=4$$

5y^{2} ni olish uchun 4y^{2} va y^{2} ni birlashtirish.

$$5y^{2}+12y+9=4$$

Ikkala tarafdan 4 ni ayirish.

$$5y^{2}+12y+9-4=0$$

5 olish uchun 9 dan 4 ni ayirish.

$$5y^{2}+12y+5=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 5 ni a, 12 ni b va 5 ni c bilan almashtiring.

$$y=\frac{-12±\sqrt{12^{2}-4\times 5\times 5}}{2\times 5}$$

12 kvadratini chiqarish.

$$y=\frac{-12±\sqrt{144-4\times 5\times 5}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$y=\frac{-12±\sqrt{144-20\times 5}}{2\times 5}$$

-20 ni 5 marotabaga ko'paytirish.

$$y=\frac{-12±\sqrt{144-100}}{2\times 5}$$

144 ni -100 ga qo'shish.

$$y=\frac{-12±\sqrt{44}}{2\times 5}$$

44 ning kvadrat ildizini chiqarish.

$$y=\frac{-12±2\sqrt{11}}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$y=\frac{-12±2\sqrt{11}}{10}$$

y=\frac{-12±2\sqrt{11}}{10} tenglamasini yeching, bunda ± musbat. -12 ni 2\sqrt{11} ga qo'shish.

$$y=\frac{2\sqrt{11}-12}{10}$$

-12+2\sqrt{11} ni 10 ga bo'lish.

$$y=\frac{\sqrt{11}-6}{5}$$

y=\frac{-12±2\sqrt{11}}{10} tenglamasini yeching, bunda ± manfiy. -12 dan 2\sqrt{11} ni ayirish.

$$y=\frac{-2\sqrt{11}-12}{10}$$

-12-2\sqrt{11} ni 10 ga bo'lish.

$$y=\frac{-\sqrt{11}-6}{5}$$

Tenglama yechildi.

$$y=\frac{\sqrt{11}-6}{5} y=\frac{-\sqrt{11}-6}{5}$$

Kvadratni yakunlash bo'yicha qadamlar

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

$$4y^{2}+12y+9+y^{2}=4$$

5y^{2} ni olish uchun 4y^{2} va y^{2} ni birlashtirish.

$$5y^{2}+12y+9=4$$

Ikkala tarafdan 9 ni ayirish.

$$5y^{2}+12y=4-9$$

-5 olish uchun 4 dan 9 ni ayirish.

$$5y^{2}+12y=-5$$

Ikki tarafini 5 ga bo‘ling.

$$\frac{5y^{2}+12y}{5}=-\frac{5}{5}$$

5 ga bo'lish 5 ga ko'paytirishni bekor qiladi.

$$y^{2}+\frac{12}{5}y=-\frac{5}{5}$$

-5 ni 5 ga bo'lish.

$$y^{2}+\frac{12}{5}y=-1$$

\frac{12}{5} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{6}{5} olish uchun. Keyin, \frac{6}{5} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$y^{2}+\frac{12}{5}y+\left(\frac{6}{5}\right)^{2}=-1+\left(\frac{6}{5}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{6}{5} kvadratini chiqarish.

$$y^{2}+\frac{12}{5}y+\frac{36}{25}=-1+\frac{36}{25}$$

-1 ni \frac{36}{25} ga qo'shish.

$$y^{2}+\frac{12}{5}y+\frac{36}{25}=\frac{11}{25}$$

y^{2}+\frac{12}{5}y+\frac{36}{25} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(y+\frac{6}{5}\right)^{2}=\frac{11}{25}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(y+\frac{6}{5}\right)^{2}}=\sqrt{\frac{11}{25}}$$

Qisqartirish.

$$y+\frac{6}{5}=\frac{\sqrt{11}}{5} y+\frac{6}{5}=-\frac{\sqrt{11}}{5}$$

Tenglamaning ikkala tarafidan \frac{6}{5} ni ayirish.

$$y=\frac{\sqrt{11}-6}{5} y=\frac{-\sqrt{11}-6}{5}$$