y^2-y+7=0 eching

$$y ^ { 2 } - y + 7 = 0$$

y uchun yechish

$$y=\frac{1+3\sqrt{3}i}{2}\approx 0,5+2,598076211i$$

$$y=\frac{-3\sqrt{3}i+1}{2}\approx 0,5-2,598076211i$$

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.

$$y^{2}-y+7=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, -1 ni b va 7 ni c bilan almashtiring.

$$y=\frac{-\left(-1\right)±\sqrt{1-4\times 7}}{2}$$

-4 ni 7 marotabaga ko'paytirish.

$$y=\frac{-\left(-1\right)±\sqrt{1-28}}{2}$$

1 ni -28 ga qo'shish.

$$y=\frac{-\left(-1\right)±\sqrt{-27}}{2}$$

-27 ning kvadrat ildizini chiqarish.

$$y=\frac{-\left(-1\right)±3\sqrt{3}i}{2}$$

-1 ning teskarisi 1 ga teng.

$$y=\frac{1±3\sqrt{3}i}{2}$$

y=\frac{1±3\sqrt{3}i}{2} tenglamasini yeching, bunda ± musbat. 1 ni 3i\sqrt{3} ga qo'shish.

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

y=\frac{1±3\sqrt{3}i}{2} tenglamasini yeching, bunda ± manfiy. 1 dan 3i\sqrt{3} ni ayirish.

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

Tenglama yechildi.

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

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.

$$y^{2}-y+7=0$$

Tenglamaning ikkala tarafidan 7 ni ayirish.

$$y^{2}-y+7-7=-7$$

O‘zidan 7 ayirilsa 0 qoladi.

$$y^{2}-y=-7$$

-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.

$$y^{2}-y+\left(-\frac{1}{2}\right)^{2}=-7+\left(-\frac{1}{2}\right)^{2}$$

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

$$y^{2}-y+\frac{1}{4}=-7+\frac{1}{4}$$

-7 ni \frac{1}{4} ga qo'shish.

$$y^{2}-y+\frac{1}{4}=-\frac{27}{4}$$

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.

$$\left(y-\frac{1}{2}\right)^{2}=-\frac{27}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(y-\frac{1}{2}\right)^{2}}=\sqrt{-\frac{27}{4}}$$

Qisqartirish.

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

\frac{1}{2} ni tenglamaning ikkala tarafiga qo'shish.

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