2x^2+3/8x+16=0 eching

$$2 x ^ { 2 } + \frac { 3 } { 8 } x + 16 = 0$$

x uchun yechish (complex solution)

$$x=\frac{-3+7\sqrt{167}i}{32}\approx -0,09375+2,826872996i$$

$$x=\frac{-7\sqrt{167}i-3}{32}\approx -0,09375-2,826872996i$$

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.

$$2x^{2}+\frac{3}{8}x+16=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 2 ni a, \frac{3}{8} ni b va 16 ni c bilan almashtiring.

$$x=\frac{-\frac{3}{8}±\sqrt{\left(\frac{3}{8}\right)^{2}-4\times 2\times 16}}{2\times 2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{3}{8} kvadratini chiqarish.

$$x=\frac{-\frac{3}{8}±\sqrt{\frac{9}{64}-4\times 2\times 16}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-\frac{3}{8}±\sqrt{\frac{9}{64}-8\times 16}}{2\times 2}$$

-8 ni 16 marotabaga ko'paytirish.

$$x=\frac{-\frac{3}{8}±\sqrt{\frac{9}{64}-128}}{2\times 2}$$

\frac{9}{64} ni -128 ga qo'shish.

$$x=\frac{-\frac{3}{8}±\sqrt{-\frac{8183}{64}}}{2\times 2}$$

-\frac{8183}{64} ning kvadrat ildizini chiqarish.

$$x=\frac{-\frac{3}{8}±\frac{7\sqrt{167}i}{8}}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{-\frac{3}{8}±\frac{7\sqrt{167}i}{8}}{4}$$

x=\frac{-\frac{3}{8}±\frac{7\sqrt{167}i}{8}}{4} tenglamasini yeching, bunda ± musbat. -\frac{3}{8} ni \frac{7i\sqrt{167}}{8} ga qo'shish.

$$x=\frac{-3+7\sqrt{167}i}{4\times 8}$$

\frac{-3+7i\sqrt{167}}{8} ni 4 ga bo'lish.

$$x=\frac{-3+7\sqrt{167}i}{32}$$

x=\frac{-\frac{3}{8}±\frac{7\sqrt{167}i}{8}}{4} tenglamasini yeching, bunda ± manfiy. -\frac{3}{8} dan \frac{7i\sqrt{167}}{8} ni ayirish.

$$x=\frac{-7\sqrt{167}i-3}{4\times 8}$$

\frac{-3-7i\sqrt{167}}{8} ni 4 ga bo'lish.

$$x=\frac{-7\sqrt{167}i-3}{32}$$

Tenglama yechildi.

$$x=\frac{-3+7\sqrt{167}i}{32} x=\frac{-7\sqrt{167}i-3}{32}$$

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.

$$2x^{2}+\frac{3}{8}x+16=0$$

Tenglamaning ikkala tarafidan 16 ni ayirish.

$$2x^{2}+\frac{3}{8}x+16-16=-16$$

O‘zidan 16 ayirilsa 0 qoladi.

$$2x^{2}+\frac{3}{8}x=-16$$

Ikki tarafini 2 ga bo‘ling.

$$\frac{2x^{2}+\frac{3}{8}x}{2}=-\frac{16}{2}$$

2 ga bo'lish 2 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{\frac{3}{8}}{2}x=-\frac{16}{2}$$

\frac{3}{8} ni 2 ga bo'lish.

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

-16 ni 2 ga bo'lish.

$$x^{2}+\frac{3}{16}x=-8$$

\frac{3}{16} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{3}{32} olish uchun. Keyin, \frac{3}{32} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{3}{16}x+\left(\frac{3}{32}\right)^{2}=-8+\left(\frac{3}{32}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{3}{32} kvadratini chiqarish.

$$x^{2}+\frac{3}{16}x+\frac{9}{1024}=-8+\frac{9}{1024}$$

-8 ni \frac{9}{1024} ga qo'shish.

$$x^{2}+\frac{3}{16}x+\frac{9}{1024}=-\frac{8183}{1024}$$

x^{2}+\frac{3}{16}x+\frac{9}{1024} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{3}{32}\right)^{2}=-\frac{8183}{1024}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{3}{32}\right)^{2}}=\sqrt{-\frac{8183}{1024}}$$

Qisqartirish.

$$x+\frac{3}{32}=\frac{7\sqrt{167}i}{32} x+\frac{3}{32}=-\frac{7\sqrt{167}i}{32}$$

Tenglamaning ikkala tarafidan \frac{3}{32} ni ayirish.

$$x=\frac{-3+7\sqrt{167}i}{32} x=\frac{-7\sqrt{167}i-3}{32}$$