1/4-y=1/4+1/y+2 eching

$$\frac { 1 } { 4 - y } = \frac { 1 } { 4 } + \frac { 1 } { y + 2 }$$

y uchun yechish

$$y=-8$$

$$y=2$$

Kvadrat tenglamalaridan foydalanish qoidalari

y qiymati -2,4 qiymatlaridan birortasiga teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini 4\left(y-4\right)\left(y+2\right) ga, 4-y,4,y+2 ning eng kichik karralisiga ko‘paytiring.

$$-8-4y=4\left(y-4\right)\left(y+2\right)\times \frac{1}{4}+4y-16$$

1 hosil qilish uchun 4 va \frac{1}{4} ni ko'paytirish.

$$-8-4y=\left(y-4\right)\left(y+2\right)+4y-16$$

y-4 ga y+2 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$-8-4y=y^{2}-2y-8+4y-16$$

2y ni olish uchun -2y va 4y ni birlashtirish.

$$-8-4y=y^{2}+2y-8-16$$

-24 olish uchun -8 dan 16 ni ayirish.

$$-8-4y=y^{2}+2y-24$$

Ikkala tarafdan y^{2} ni ayirish.

$$-8-4y-y^{2}=2y-24$$

Ikkala tarafdan 2y ni ayirish.

$$-8-4y-y^{2}-2y=-24$$

-6y ni olish uchun -4y va -2y ni birlashtirish.

$$-8-6y-y^{2}=-24$$

24 ni ikki tarafga qo’shing.

$$-8-6y-y^{2}+24=0$$

16 olish uchun -8 va 24'ni qo'shing.

$$16-6y-y^{2}=0$$

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni kvadrat formulasi bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat formula ikki yechmni taqdim qiladi, biri ± qo'shish bo'lganda, va ikkinchisi ayiruv bo'lganda.

$$-y^{2}-6y+16=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} -1 ni a, -6 ni b va 16 ni c bilan almashtiring.

$$y=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\left(-1\right)\times 16}}{2\left(-1\right)}$$

-6 kvadratini chiqarish.

$$y=\frac{-\left(-6\right)±\sqrt{36-4\left(-1\right)\times 16}}{2\left(-1\right)}$$

-4 ni -1 marotabaga ko'paytirish.

$$y=\frac{-\left(-6\right)±\sqrt{36+4\times 16}}{2\left(-1\right)}$$

4 ni 16 marotabaga ko'paytirish.

$$y=\frac{-\left(-6\right)±\sqrt{36+64}}{2\left(-1\right)}$$

36 ni 64 ga qo'shish.

$$y=\frac{-\left(-6\right)±\sqrt{100}}{2\left(-1\right)}$$

100 ning kvadrat ildizini chiqarish.

$$y=\frac{-\left(-6\right)±10}{2\left(-1\right)}$$

-6 ning teskarisi 6 ga teng.

$$y=\frac{6±10}{2\left(-1\right)}$$

2 ni -1 marotabaga ko'paytirish.

$$y=\frac{6±10}{-2}$$

y=\frac{6±10}{-2} tenglamasini yeching, bunda ± musbat. 6 ni 10 ga qo'shish.

$$y=\frac{16}{-2}$$

16 ni -2 ga bo'lish.

$$y=-8$$

y=\frac{6±10}{-2} tenglamasini yeching, bunda ± manfiy. 6 dan 10 ni ayirish.

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

-4 ni -2 ga bo'lish.

$$y=2$$

Tenglama yechildi.

$$y=-8 y=2$$

Kvadratni yakunlash bo'yicha qadamlar

y qiymati -2,4 qiymatlaridan birortasiga teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini 4\left(y-4\right)\left(y+2\right) ga, 4-y,4,y+2 ning eng kichik karralisiga ko‘paytiring.

$$-8-4y=4\left(y-4\right)\left(y+2\right)\times \frac{1}{4}+4y-16$$

1 hosil qilish uchun 4 va \frac{1}{4} ni ko'paytirish.

$$-8-4y=\left(y-4\right)\left(y+2\right)+4y-16$$

y-4 ga y+2 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$-8-4y=y^{2}-2y-8+4y-16$$

2y ni olish uchun -2y va 4y ni birlashtirish.

$$-8-4y=y^{2}+2y-8-16$$

-24 olish uchun -8 dan 16 ni ayirish.

$$-8-4y=y^{2}+2y-24$$

Ikkala tarafdan y^{2} ni ayirish.

$$-8-4y-y^{2}=2y-24$$

Ikkala tarafdan 2y ni ayirish.

$$-8-4y-y^{2}-2y=-24$$

-6y ni olish uchun -4y va -2y ni birlashtirish.

$$-8-6y-y^{2}=-24$$

8 ni ikki tarafga qo’shing.

$$-6y-y^{2}=-24+8$$

-16 olish uchun -24 va 8'ni qo'shing.

$$-6y-y^{2}=-16$$

Bu kabi kvadrat tenglamalarni kvadratni yakunlab yechish mumkin. Kvadratni yechish uchun tenglama avval ushbu shaklda bo'lishi shart: x^{2}+bx=c.

$$-y^{2}-6y=-16$$

Ikki tarafini -1 ga bo‘ling.

$$\frac{-y^{2}-6y}{-1}=-\frac{16}{-1}$$

-1 ga bo'lish -1 ga ko'paytirishni bekor qiladi.

$$y^{2}+\left(-\frac{6}{-1}\right)y=-\frac{16}{-1}$$

-6 ni -1 ga bo'lish.

$$y^{2}+6y=-\frac{16}{-1}$$

-16 ni -1 ga bo'lish.

$$y^{2}+6y=16$$

6 ni bo‘lish, x shartining koeffitsienti, 2 ga 3 olish uchun. Keyin, 3 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$y^{2}+6y+3^{2}=16+3^{2}$$

3 kvadratini chiqarish.

$$y^{2}+6y+9=16+9$$

16 ni 9 ga qo'shish.

$$y^{2}+6y+9=25$$

y^{2}+6y+9 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(y+3\right)^{2}=25$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(y+3\right)^{2}}=\sqrt{25}$$

Qisqartirish.

$$y+3=5 y+3=-5$$

Tenglamaning ikkala tarafidan 3 ni ayirish.

$$y=2 y=-8$$