-4a^2-5a+1=0 eching

$$- 4 a ^ { 2 } - 5 a + 1 = 0$$

a uchun yechish

$$a=\frac{\sqrt{41}-5}{8}\approx 0,17539053$$

$$a=\frac{-\sqrt{41}-5}{8}\approx -1,42539053$$

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.

$$-4a^{2}-5a+1=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} -4 ni a, -5 ni b va 1 ni c bilan almashtiring.

$$a=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-4\right)}}{2\left(-4\right)}$$

-5 kvadratini chiqarish.

$$a=\frac{-\left(-5\right)±\sqrt{25-4\left(-4\right)}}{2\left(-4\right)}$$

-4 ni -4 marotabaga ko'paytirish.

$$a=\frac{-\left(-5\right)±\sqrt{25+16}}{2\left(-4\right)}$$

25 ni 16 ga qo'shish.

$$a=\frac{-\left(-5\right)±\sqrt{41}}{2\left(-4\right)}$$

-5 ning teskarisi 5 ga teng.

$$a=\frac{5±\sqrt{41}}{2\left(-4\right)}$$

2 ni -4 marotabaga ko'paytirish.

$$a=\frac{5±\sqrt{41}}{-8}$$

a=\frac{5±\sqrt{41}}{-8} tenglamasini yeching, bunda ± musbat. 5 ni \sqrt{41} ga qo'shish.

$$a=\frac{\sqrt{41}+5}{-8}$$

5+\sqrt{41} ni -8 ga bo'lish.

$$a=\frac{-\sqrt{41}-5}{8}$$

a=\frac{5±\sqrt{41}}{-8} tenglamasini yeching, bunda ± manfiy. 5 dan \sqrt{41} ni ayirish.

$$a=\frac{5-\sqrt{41}}{-8}$$

5-\sqrt{41} ni -8 ga bo'lish.

$$a=\frac{\sqrt{41}-5}{8}$$

Tenglama yechildi.

$$a=\frac{-\sqrt{41}-5}{8} a=\frac{\sqrt{41}-5}{8}$$

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.

$$-4a^{2}-5a+1=0$$

Tenglamaning ikkala tarafidan 1 ni ayirish.

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

O‘zidan 1 ayirilsa 0 qoladi.

$$-4a^{2}-5a=-1$$

Ikki tarafini -4 ga bo‘ling.

$$\frac{-4a^{2}-5a}{-4}=-\frac{1}{-4}$$

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

$$a^{2}+\left(-\frac{5}{-4}\right)a=-\frac{1}{-4}$$

-5 ni -4 ga bo'lish.

$$a^{2}+\frac{5}{4}a=-\frac{1}{-4}$$

-1 ni -4 ga bo'lish.

$$a^{2}+\frac{5}{4}a=\frac{1}{4}$$

\frac{5}{4} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{5}{8} olish uchun. Keyin, \frac{5}{8} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$a^{2}+\frac{5}{4}a+\left(\frac{5}{8}\right)^{2}=\frac{1}{4}+\left(\frac{5}{8}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{5}{8} kvadratini chiqarish.

$$a^{2}+\frac{5}{4}a+\frac{25}{64}=\frac{1}{4}+\frac{25}{64}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{1}{4} ni \frac{25}{64} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$a^{2}+\frac{5}{4}a+\frac{25}{64}=\frac{41}{64}$$

a^{2}+\frac{5}{4}a+\frac{25}{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.

$$\left(a+\frac{5}{8}\right)^{2}=\frac{41}{64}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(a+\frac{5}{8}\right)^{2}}=\sqrt{\frac{41}{64}}$$

Qisqartirish.

$$a+\frac{5}{8}=\frac{\sqrt{41}}{8} a+\frac{5}{8}=-\frac{\sqrt{41}}{8}$$

Tenglamaning ikkala tarafidan \frac{5}{8} ni ayirish.

$$a=\frac{\sqrt{41}-5}{8} a=\frac{-\sqrt{41}-5}{8}$$