4x^2+6x+10=0 eching

$$4 x ^ { 2 } + 6 x + 10 = 0$$

x uchun yechish (complex solution)

$$x=\frac{-3+\sqrt{31}i}{4}\approx -0,75+1,391941091i$$

$$x=\frac{-\sqrt{31}i-3}{4}\approx -0,75-1,391941091i$$

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.

$$4x^{2}+6x+10=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 4 ni a, 6 ni b va 10 ni c bilan almashtiring.

$$x=\frac{-6±\sqrt{6^{2}-4\times 4\times 10}}{2\times 4}$$

6 kvadratini chiqarish.

$$x=\frac{-6±\sqrt{36-4\times 4\times 10}}{2\times 4}$$

-4 ni 4 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36-16\times 10}}{2\times 4}$$

-16 ni 10 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36-160}}{2\times 4}$$

36 ni -160 ga qo'shish.

$$x=\frac{-6±\sqrt{-124}}{2\times 4}$$

-124 ning kvadrat ildizini chiqarish.

$$x=\frac{-6±2\sqrt{31}i}{2\times 4}$$

2 ni 4 marotabaga ko'paytirish.

$$x=\frac{-6±2\sqrt{31}i}{8}$$

x=\frac{-6±2\sqrt{31}i}{8} tenglamasini yeching, bunda ± musbat. -6 ni 2i\sqrt{31} ga qo'shish.

$$x=\frac{-6+2\sqrt{31}i}{8}$$

-6+2i\sqrt{31} ni 8 ga bo'lish.

$$x=\frac{-3+\sqrt{31}i}{4}$$

x=\frac{-6±2\sqrt{31}i}{8} tenglamasini yeching, bunda ± manfiy. -6 dan 2i\sqrt{31} ni ayirish.

$$x=\frac{-2\sqrt{31}i-6}{8}$$

-6-2i\sqrt{31} ni 8 ga bo'lish.

$$x=\frac{-\sqrt{31}i-3}{4}$$

Tenglama yechildi.

$$x=\frac{-3+\sqrt{31}i}{4} x=\frac{-\sqrt{31}i-3}{4}$$

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.

$$4x^{2}+6x+10=0$$

Tenglamaning ikkala tarafidan 10 ni ayirish.

$$4x^{2}+6x+10-10=-10$$

O‘zidan 10 ayirilsa 0 qoladi.

$$4x^{2}+6x=-10$$

Ikki tarafini 4 ga bo‘ling.

$$\frac{4x^{2}+6x}{4}=-\frac{10}{4}$$

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

$$x^{2}+\frac{6}{4}x=-\frac{10}{4}$$

\frac{6}{4} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}+\frac{3}{2}x=-\frac{10}{4}$$

\frac{-10}{4} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}+\frac{3}{2}x=-\frac{5}{2}$$

\frac{3}{2} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{3}{4} olish uchun. Keyin, \frac{3}{4} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{3}{2}x+\left(\frac{3}{4}\right)^{2}=-\frac{5}{2}+\left(\frac{3}{4}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{3}{4} kvadratini chiqarish.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=-\frac{5}{2}+\frac{9}{16}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{5}{2} ni \frac{9}{16} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=-\frac{31}{16}$$

x^{2}+\frac{3}{2}x+\frac{9}{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.

$$\left(x+\frac{3}{4}\right)^{2}=-\frac{31}{16}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{3}{4}\right)^{2}}=\sqrt{-\frac{31}{16}}$$

Qisqartirish.

$$x+\frac{3}{4}=\frac{\sqrt{31}i}{4} x+\frac{3}{4}=-\frac{\sqrt{31}i}{4}$$

Tenglamaning ikkala tarafidan \frac{3}{4} ni ayirish.

$$x=\frac{-3+\sqrt{31}i}{4} x=\frac{-\sqrt{31}i-3}{4}$$