4x^2+4x+9=0 eching

$$4 x ^ { 2 } + 4 x + 9 = 0$$

x uchun yechish (complex solution)

$$x=-\frac{1}{2}+\sqrt{2}i\approx -0,5+1,414213562i$$

$$x=-\sqrt{2}i-\frac{1}{2}\approx -0,5-1,414213562i$$

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.

$$4x^{2}+4x+9=0$$

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 9 ni c bilan almashtiring.

$$x=\frac{-4±\sqrt{4^{2}-4\times 4\times 9}}{2\times 4}$$

4 kvadratini chiqarish.

$$x=\frac{-4±\sqrt{16-4\times 4\times 9}}{2\times 4}$$

-4 ni 4 marotabaga ko'paytirish.

$$x=\frac{-4±\sqrt{16-16\times 9}}{2\times 4}$$

-16 ni 9 marotabaga ko'paytirish.

$$x=\frac{-4±\sqrt{16-144}}{2\times 4}$$

16 ni -144 ga qo'shish.

$$x=\frac{-4±\sqrt{-128}}{2\times 4}$$

-128 ning kvadrat ildizini chiqarish.

$$x=\frac{-4±8\sqrt{2}i}{2\times 4}$$

2 ni 4 marotabaga ko'paytirish.

$$x=\frac{-4±8\sqrt{2}i}{8}$$

x=\frac{-4±8\sqrt{2}i}{8} tenglamasini yeching, bunda ± musbat. -4 ni 8i\sqrt{2} ga qo'shish.

$$x=\frac{-4+2\times 2^{\frac{5}{2}}i}{8}$$

-4+2i\times 2^{\frac{5}{2}} ni 8 ga bo'lish.

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

x=\frac{-4±8\sqrt{2}i}{8} tenglamasini yeching, bunda ± manfiy. -4 dan 8i\sqrt{2} ni ayirish.

$$x=\frac{-2\times 2^{\frac{5}{2}}i-4}{8}$$

-4-2i\times 2^{\frac{5}{2}} ni 8 ga bo'lish.

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

Tenglama yechildi.

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

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.

$$4x^{2}+4x+9=0$$

Tenglamaning ikkala tarafidan 9 ni ayirish.

$$4x^{2}+4x+9-9=-9$$

O‘zidan 9 ayirilsa 0 qoladi.

$$4x^{2}+4x=-9$$

Ikki tarafini 4 ga bo‘ling.

$$\frac{4x^{2}+4x}{4}=-\frac{9}{4}$$

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

$$x^{2}+\frac{4}{4}x=-\frac{9}{4}$$

4 ni 4 ga bo'lish.

$$x^{2}+x=-\frac{9}{4}$$

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.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=-\frac{9}{4}+\left(\frac{1}{2}\right)^{2}$$

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

$$x^{2}+x+\frac{1}{4}=\frac{-9+1}{4}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{9}{4} ni \frac{1}{4} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

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

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.

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

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

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

Qisqartirish.

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

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

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