2x+1=4x^2+4x+5 eching

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

x uchun yechish (complex solution)

$$x=\frac{-1+\sqrt{15}i}{4}\approx -0,25+0,968245837i$$

$$x=\frac{-\sqrt{15}i-1}{4}\approx -0,25-0,968245837i$$

Kvadrat tenglamalaridan foydalanish qoidalari

Ikkala tarafdan 4x^{2} ni ayirish.

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

Ikkala tarafdan 4x ni ayirish.

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

-2x ni olish uchun 2x va -4x ni birlashtirish.

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

Ikkala tarafdan 5 ni ayirish.

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

-4 olish uchun 1 dan 5 ni ayirish.

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

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.

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

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

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

-2 kvadratini chiqarish.

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

-4 ni -4 marotabaga ko'paytirish.

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

16 ni -4 marotabaga ko'paytirish.

$$x=\frac{-\left(-2\right)±\sqrt{4-64}}{2\left(-4\right)}$$

4 ni -64 ga qo'shish.

$$x=\frac{-\left(-2\right)±\sqrt{-60}}{2\left(-4\right)}$$

-60 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-2\right)±2\sqrt{15}i}{2\left(-4\right)}$$

-2 ning teskarisi 2 ga teng.

$$x=\frac{2±2\sqrt{15}i}{2\left(-4\right)}$$

2 ni -4 marotabaga ko'paytirish.

$$x=\frac{2±2\sqrt{15}i}{-8}$$

x=\frac{2±2\sqrt{15}i}{-8} tenglamasini yeching, bunda ± musbat. 2 ni 2i\sqrt{15} ga qo'shish.

$$x=\frac{2+2\sqrt{15}i}{-8}$$

2+2i\sqrt{15} ni -8 ga bo'lish.

$$x=\frac{-\sqrt{15}i-1}{4}$$

x=\frac{2±2\sqrt{15}i}{-8} tenglamasini yeching, bunda ± manfiy. 2 dan 2i\sqrt{15} ni ayirish.

$$x=\frac{-2\sqrt{15}i+2}{-8}$$

2-2i\sqrt{15} ni -8 ga bo'lish.

$$x=\frac{-1+\sqrt{15}i}{4}$$

Tenglama yechildi.

$$x=\frac{-\sqrt{15}i-1}{4} x=\frac{-1+\sqrt{15}i}{4}$$

Kvadratni yakunlash bo'yicha qadamlar

Ikkala tarafdan 4x^{2} ni ayirish.

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

Ikkala tarafdan 4x ni ayirish.

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

-2x ni olish uchun 2x va -4x ni birlashtirish.

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

Ikkala tarafdan 1 ni ayirish.

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

4 olish uchun 5 dan 1 ni ayirish.

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

Bu kabi kvadrat tenglamalarni kvadratni yakunlab yechish mumkin. Kvadratni yechish uchun tenglama avval ushbu shaklda bo'lishi shart: x^{2}+bx=c.

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

Ikki tarafini -4 ga bo‘ling.

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

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

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

\frac{-2}{-4} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

4 ni -4 ga bo'lish.

$$x^{2}+\frac{1}{2}x=-1$$

\frac{1}{2} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{1}{4} olish uchun. Keyin, \frac{1}{4} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{1}{4} kvadratini chiqarish.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=-1+\frac{1}{16}$$

-1 ni \frac{1}{16} ga qo'shish.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=-\frac{15}{16}$$

x^{2}+\frac{1}{2}x+\frac{1}{16} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{1}{4}\right)^{2}=-\frac{15}{16}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{-\frac{15}{16}}$$

Qisqartirish.

$$x+\frac{1}{4}=\frac{\sqrt{15}i}{4} x+\frac{1}{4}=-\frac{\sqrt{15}i}{4}$$

Tenglamaning ikkala tarafidan \frac{1}{4} ni ayirish.

$$x=\frac{-1+\sqrt{15}i}{4} x=\frac{-\sqrt{15}i-1}{4}$$