4x^2+28x+53=0 eching

$$4 x ^ { 2 } + 28 x + 53 = 0$$

x uchun yechish (complex solution)

$$x=-\frac{7}{2}+i=-3,5+i$$

$$x=-\frac{7}{2}-i=-3,5-i$$

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.

$$4x^{2}+28x+53=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 4 ni a, 28 ni b va 53 ni c bilan almashtiring.

$$x=\frac{-28±\sqrt{28^{2}-4\times 4\times 53}}{2\times 4}$$

28 kvadratini chiqarish.

$$x=\frac{-28±\sqrt{784-4\times 4\times 53}}{2\times 4}$$

-4 ni 4 marotabaga ko'paytirish.

$$x=\frac{-28±\sqrt{784-16\times 53}}{2\times 4}$$

-16 ni 53 marotabaga ko'paytirish.

$$x=\frac{-28±\sqrt{784-848}}{2\times 4}$$

784 ni -848 ga qo'shish.

$$x=\frac{-28±\sqrt{-64}}{2\times 4}$$

-64 ning kvadrat ildizini chiqarish.

$$x=\frac{-28±8i}{2\times 4}$$

2 ni 4 marotabaga ko'paytirish.

$$x=\frac{-28±8i}{8}$$

x=\frac{-28±8i}{8} tenglamasini yeching, bunda ± musbat. -28 ni 8i ga qo'shish.

$$x=\frac{-28+8i}{8}$$

-28+8i ni 8 ga bo'lish.

$$x=-\frac{7}{2}+i$$

x=\frac{-28±8i}{8} tenglamasini yeching, bunda ± manfiy. -28 dan 8i ni ayirish.

$$x=\frac{-28-8i}{8}$$

-28-8i ni 8 ga bo'lish.

$$x=-\frac{7}{2}-i$$

Tenglama yechildi.

$$x=-\frac{7}{2}+i x=-\frac{7}{2}-i$$

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.

$$4x^{2}+28x+53=0$$

Tenglamaning ikkala tarafidan 53 ni ayirish.

$$4x^{2}+28x+53-53=-53$$

O‘zidan 53 ayirilsa 0 qoladi.

$$4x^{2}+28x=-53$$

Ikki tarafini 4 ga bo‘ling.

$$\frac{4x^{2}+28x}{4}=-\frac{53}{4}$$

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

$$x^{2}+\frac{28}{4}x=-\frac{53}{4}$$

28 ni 4 ga bo'lish.

$$x^{2}+7x=-\frac{53}{4}$$

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

$$x^{2}+7x+\left(\frac{7}{2}\right)^{2}=-\frac{53}{4}+\left(\frac{7}{2}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{7}{2} kvadratini chiqarish.

$$x^{2}+7x+\frac{49}{4}=\frac{-53+49}{4}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{53}{4} ni \frac{49}{4} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

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

x^{2}+7x+\frac{49}{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.

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

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{7}{2}\right)^{2}}=\sqrt{-1}$$

Qisqartirish.

$$x+\frac{7}{2}=i x+\frac{7}{2}=-i$$

Tenglamaning ikkala tarafidan \frac{7}{2} ni ayirish.

$$x=-\frac{7}{2}+i x=-\frac{7}{2}-i$$