4x^2+4x=5 eching

$$4 x ^ { 2 } + 4 x = 5$$

x uchun yechish

$$x=\frac{\sqrt{6}-1}{2}\approx 0,724744871$$

$$x=\frac{-\sqrt{6}-1}{2}\approx -1,724744871$$

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.

$$4x^{2}+4x=5$$

Tenglamaning ikkala tarafidan 5 ni ayirish.

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

O‘zidan 5 ayirilsa 0 qoladi.

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

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

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

4 kvadratini chiqarish.

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

-4 ni 4 marotabaga ko'paytirish.

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

-16 ni -5 marotabaga ko'paytirish.

$$x=\frac{-4±\sqrt{16+80}}{2\times 4}$$

16 ni 80 ga qo'shish.

$$x=\frac{-4±\sqrt{96}}{2\times 4}$$

96 ning kvadrat ildizini chiqarish.

$$x=\frac{-4±4\sqrt{6}}{2\times 4}$$

2 ni 4 marotabaga ko'paytirish.

$$x=\frac{-4±4\sqrt{6}}{8}$$

x=\frac{-4±4\sqrt{6}}{8} tenglamasini yeching, bunda ± musbat. -4 ni 4\sqrt{6} ga qo'shish.

$$x=\frac{4\sqrt{6}-4}{8}$$

-4+4\sqrt{6} ni 8 ga bo'lish.

$$x=\frac{\sqrt{6}-1}{2}$$

x=\frac{-4±4\sqrt{6}}{8} tenglamasini yeching, bunda ± manfiy. -4 dan 4\sqrt{6} ni ayirish.

$$x=\frac{-4\sqrt{6}-4}{8}$$

-4-4\sqrt{6} ni 8 ga bo'lish.

$$x=\frac{-\sqrt{6}-1}{2}$$

Tenglama yechildi.

$$x=\frac{\sqrt{6}-1}{2} x=\frac{-\sqrt{6}-1}{2}$$

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.

$$4x^{2}+4x=5$$

Ikki tarafini 4 ga bo‘ling.

$$\frac{4x^{2}+4x}{4}=\frac{5}{4}$$

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

$$x^{2}+\frac{4}{4}x=\frac{5}{4}$$

4 ni 4 ga bo'lish.

$$x^{2}+x=\frac{5}{4}$$

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.

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

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

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

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{5}{4} ni \frac{1}{4} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

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

x^{2}+x+\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.

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

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

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

Qisqartirish.

$$x+\frac{1}{2}=\frac{\sqrt{6}}{2} x+\frac{1}{2}=-\frac{\sqrt{6}}{2}$$

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

$$x=\frac{\sqrt{6}-1}{2} x=\frac{-\sqrt{6}-1}{2}$$