4y^2-7y+1=0 eching

$$4 y ^ { 2 } - 7 y + 1 = 0$$

y uchun yechish

$$y = \frac{\sqrt{33} + 7}{8} \approx 1,593070331$$

$$y=\frac{7-\sqrt{33}}{8}\approx 0,156929669$$

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.

$$4y^{2}-7y+1=0$$

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

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

-7 kvadratini chiqarish.

$$y=\frac{-\left(-7\right)±\sqrt{49-4\times 4}}{2\times 4}$$

-4 ni 4 marotabaga ko'paytirish.

$$y=\frac{-\left(-7\right)±\sqrt{49-16}}{2\times 4}$$

49 ni -16 ga qo'shish.

$$y=\frac{-\left(-7\right)±\sqrt{33}}{2\times 4}$$

-7 ning teskarisi 7 ga teng.

$$y=\frac{7±\sqrt{33}}{2\times 4}$$

2 ni 4 marotabaga ko'paytirish.

$$y=\frac{7±\sqrt{33}}{8}$$

y=\frac{7±\sqrt{33}}{8} tenglamasini yeching, bunda ± musbat. 7 ni \sqrt{33} ga qo'shish.

$$y=\frac{\sqrt{33}+7}{8}$$

y=\frac{7±\sqrt{33}}{8} tenglamasini yeching, bunda ± manfiy. 7 dan \sqrt{33} ni ayirish.

$$y=\frac{7-\sqrt{33}}{8}$$

Tenglama yechildi.

$$y=\frac{\sqrt{33}+7}{8} y=\frac{7-\sqrt{33}}{8}$$

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.

$$4y^{2}-7y+1=0$$

Tenglamaning ikkala tarafidan 1 ni ayirish.

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

O‘zidan 1 ayirilsa 0 qoladi.

$$4y^{2}-7y=-1$$

Ikki tarafini 4 ga bo‘ling.

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

4 ga bo'lish 4 ga ko'paytirishni bekor qiladi.

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

-\frac{7}{4} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{7}{8} olish uchun. Keyin, -\frac{7}{8} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$y^{2}-\frac{7}{4}y+\left(-\frac{7}{8}\right)^{2}=-\frac{1}{4}+\left(-\frac{7}{8}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{7}{8} kvadratini chiqarish.

$$y^{2}-\frac{7}{4}y+\frac{49}{64}=-\frac{1}{4}+\frac{49}{64}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{1}{4} ni \frac{49}{64} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$y^{2}-\frac{7}{4}y+\frac{49}{64}=\frac{33}{64}$$

y^{2}-\frac{7}{4}y+\frac{49}{64} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(y-\frac{7}{8}\right)^{2}=\frac{33}{64}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(y-\frac{7}{8}\right)^{2}}=\sqrt{\frac{33}{64}}$$

Qisqartirish.

$$y-\frac{7}{8}=\frac{\sqrt{33}}{8} y-\frac{7}{8}=-\frac{\sqrt{33}}{8}$$

\frac{7}{8} ni tenglamaning ikkala tarafiga qo'shish.

$$y=\frac{\sqrt{33}+7}{8} y=\frac{7-\sqrt{33}}{8}$$