2y^2-4y+5y=1 eching

$$2 y ^ { 2 } - 4 y + 5 y = 1$$

y uchun yechish

$$y=-1$$

$$y=\frac{1}{2}=0,5$$

Guruhlash orqali faktorlash qadamlari

y ni olish uchun -4y va 5y ni birlashtirish.

$$2y^{2}+y=1$$

Ikkala tarafdan 1 ni ayirish.

$$2y^{2}+y-1=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon 2y^{2}+ay+by-1 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=1 ab=2\left(-1\right)=-2$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b musbat boʻlganda, musbat sonda manfiyga nisbatdan kattaroq mutlaq qiymat bor. Faqat bundan juftlik tizim yechimidir.

$$a=-1 b=2$$

2y^{2}+y-1 ni \left(2y^{2}-y\right)+\left(2y-1\right) sifatida qaytadan yozish.

$$\left(2y^{2}-y\right)+\left(2y-1\right)$$

2y^{2}-y ichida y ni ajrating.

$$y\left(2y-1\right)+2y-1$$

Distributiv funktsiyasidan foydalangan holda 2y-1 umumiy terminini chiqaring.

$$\left(2y-1\right)\left(y+1\right)$$

Tenglamani yechish uchun 2y-1=0 va y+1=0 ni yeching.

$$y=\frac{1}{2} y=-1$$

Kvadrat tenglamalaridan foydalanish qoidalari

y ni olish uchun -4y va 5y ni birlashtirish.

$$2y^{2}+y=1$$

Ikkala tarafdan 1 ni ayirish.

$$2y^{2}+y-1=0$$

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

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

1 kvadratini chiqarish.

$$y=\frac{-1±\sqrt{1-4\times 2\left(-1\right)}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$y=\frac{-1±\sqrt{1-8\left(-1\right)}}{2\times 2}$$

-8 ni -1 marotabaga ko'paytirish.

$$y=\frac{-1±\sqrt{1+8}}{2\times 2}$$

1 ni 8 ga qo'shish.

$$y=\frac{-1±\sqrt{9}}{2\times 2}$$

9 ning kvadrat ildizini chiqarish.

$$y=\frac{-1±3}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$y=\frac{-1±3}{4}$$

y=\frac{-1±3}{4} tenglamasini yeching, bunda ± musbat. -1 ni 3 ga qo'shish.

$$y=\frac{2}{4}$$

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

$$y=\frac{1}{2}$$

y=\frac{-1±3}{4} tenglamasini yeching, bunda ± manfiy. -1 dan 3 ni ayirish.

$$y=-\frac{4}{4}$$

-4 ni 4 ga bo'lish.

$$y=-1$$

Tenglama yechildi.

$$y=\frac{1}{2} y=-1$$

Kvadratni yakunlash bo'yicha qadamlar

y ni olish uchun -4y va 5y ni birlashtirish.

$$2y^{2}+y=1$$

Ikki tarafini 2 ga bo‘ling.

$$\frac{2y^{2}+y}{2}=\frac{1}{2}$$

2 ga bo'lish 2 ga ko'paytirishni bekor qiladi.

$$y^{2}+\frac{1}{2}y=\frac{1}{2}$$

\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.

$$y^{2}+\frac{1}{2}y+\left(\frac{1}{4}\right)^{2}=\frac{1}{2}+\left(\frac{1}{4}\right)^{2}$$

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

$$y^{2}+\frac{1}{2}y+\frac{1}{16}=\frac{1}{2}+\frac{1}{16}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{1}{2} ni \frac{1}{16} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$y^{2}+\frac{1}{2}y+\frac{1}{16}=\frac{9}{16}$$

y^{2}+\frac{1}{2}y+\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.

$$\left(y+\frac{1}{4}\right)^{2}=\frac{9}{16}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(y+\frac{1}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Qisqartirish.

$$y+\frac{1}{4}=\frac{3}{4} y+\frac{1}{4}=-\frac{3}{4}$$

Tenglamaning ikkala tarafidan \frac{1}{4} ni ayirish.

$$y=\frac{1}{2} y=-1$$