4x^2-5x-1=0 eching

$$4 x ^ { 2 } - 5 x - 1 = 0$$

x uchun yechish

$$x = \frac{\sqrt{41} + 5}{8} \approx 1,42539053$$

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

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.

$$4x^{2}-5x-1=0$$

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.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 4\left(-1\right)}}{2\times 4}$$

-5 kvadratini chiqarish.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 4\left(-1\right)}}{2\times 4}$$

-4 ni 4 marotabaga ko'paytirish.

$$x=\frac{-\left(-5\right)±\sqrt{25-16\left(-1\right)}}{2\times 4}$$

-16 ni -1 marotabaga ko'paytirish.

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

25 ni 16 ga qo'shish.

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

-5 ning teskarisi 5 ga teng.

$$x=\frac{5±\sqrt{41}}{2\times 4}$$

2 ni 4 marotabaga ko'paytirish.

$$x=\frac{5±\sqrt{41}}{8}$$

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

$$x=\frac{\sqrt{41}+5}{8}$$

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

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

Tenglama yechildi.

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

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.

$$4x^{2}-5x-1=0$$

1 ni tenglamaning ikkala tarafiga qo'shish.

$$4x^{2}-5x-1-\left(-1\right)=-\left(-1\right)$$

O‘zidan -1 ayirilsa 0 qoladi.

$$4x^{2}-5x=-\left(-1\right)$$

0 dan -1 ni ayirish.

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

Ikki tarafini 4 ga bo‘ling.

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

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

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

-\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.

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

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

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

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.

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

x^{2}-\frac{5}{4}x+\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.

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

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

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

Qisqartirish.

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

\frac{5}{8} ni tenglamaning ikkala tarafiga qo'shish.

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