2y^2+2y-1=0 eching

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

y uchun yechish

$$y=\frac{\sqrt{3}-1}{2}\approx 0,366025404$$

$$y=\frac{-\sqrt{3}-1}{2}\approx -1,366025404$$

Kvadrat tenglamalaridan foydalanish qoidalari

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.

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

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

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

2 kvadratini chiqarish.

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

-4 ni 2 marotabaga ko'paytirish.

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

-8 ni -1 marotabaga ko'paytirish.

$$y=\frac{-2±\sqrt{4+8}}{2\times 2}$$

4 ni 8 ga qo'shish.

$$y=\frac{-2±\sqrt{12}}{2\times 2}$$

12 ning kvadrat ildizini chiqarish.

$$y=\frac{-2±2\sqrt{3}}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$y=\frac{-2±2\sqrt{3}}{4}$$

y=\frac{-2±2\sqrt{3}}{4} tenglamasini yeching, bunda ± musbat. -2 ni 2\sqrt{3} ga qo'shish.

$$y=\frac{2\sqrt{3}-2}{4}$$

-2+2\sqrt{3} ni 4 ga bo'lish.

$$y=\frac{\sqrt{3}-1}{2}$$

y=\frac{-2±2\sqrt{3}}{4} tenglamasini yeching, bunda ± manfiy. -2 dan 2\sqrt{3} ni ayirish.

$$y=\frac{-2\sqrt{3}-2}{4}$$

-2-2\sqrt{3} ni 4 ga bo'lish.

$$y=\frac{-\sqrt{3}-1}{2}$$

Tenglama yechildi.

$$y=\frac{\sqrt{3}-1}{2} y=\frac{-\sqrt{3}-1}{2}$$

Kvadratni yakunlash bo'yicha qadamlar

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

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

1 ni tenglamaning ikkala tarafiga qo'shish.

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

O‘zidan -1 ayirilsa 0 qoladi.

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

0 dan -1 ni ayirish.

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

Ikki tarafini 2 ga bo‘ling.

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

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

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

2 ni 2 ga bo'lish.

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

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

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{1}{2} kvadratini chiqarish.

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

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{1}{2} ni \frac{1}{4} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$y^{2}+y+\frac{1}{4}=\frac{3}{4}$$

y^{2}+y+\frac{1}{4} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

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

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

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

Qisqartirish.

$$y+\frac{1}{2}=\frac{\sqrt{3}}{2} y+\frac{1}{2}=-\frac{\sqrt{3}}{2}$$

Tenglamaning ikkala tarafidan \frac{1}{2} ni ayirish.

$$y=\frac{\sqrt{3}-1}{2} y=\frac{-\sqrt{3}-1}{2}$$