2x^2+3x+17=1 eching

$$2 x ^ { 2 } + 3 x + 17 = 1$$

x uchun yechish (complex solution)

$$x=\frac{-3+\sqrt{119}i}{4}\approx -0,75+2,727178029i$$

$$x=\frac{-\sqrt{119}i-3}{4}\approx -0,75-2,727178029i$$

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.

$$2x^{2}+3x+17=1$$

Tenglamaning ikkala tarafidan 1 ni ayirish.

$$2x^{2}+3x+17-1=1-1$$

O‘zidan 1 ayirilsa 0 qoladi.

$$2x^{2}+3x+17-1=0$$

17 dan 1 ni ayirish.

$$2x^{2}+3x+16=0$$

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

$$x=\frac{-3±\sqrt{3^{2}-4\times 2\times 16}}{2\times 2}$$

3 kvadratini chiqarish.

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

-4 ni 2 marotabaga ko'paytirish.

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

-8 ni 16 marotabaga ko'paytirish.

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

9 ni -128 ga qo'shish.

$$x=\frac{-3±\sqrt{-119}}{2\times 2}$$

-119 ning kvadrat ildizini chiqarish.

$$x=\frac{-3±\sqrt{119}i}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{-3±\sqrt{119}i}{4}$$

x=\frac{-3±\sqrt{119}i}{4} tenglamasini yeching, bunda ± musbat. -3 ni i\sqrt{119} ga qo'shish.

$$x=\frac{-3+\sqrt{119}i}{4}$$

x=\frac{-3±\sqrt{119}i}{4} tenglamasini yeching, bunda ± manfiy. -3 dan i\sqrt{119} ni ayirish.

$$x=\frac{-\sqrt{119}i-3}{4}$$

Tenglama yechildi.

$$x=\frac{-3+\sqrt{119}i}{4} x=\frac{-\sqrt{119}i-3}{4}$$

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.

$$2x^{2}+3x+17=1$$

Tenglamaning ikkala tarafidan 17 ni ayirish.

$$2x^{2}+3x+17-17=1-17$$

O‘zidan 17 ayirilsa 0 qoladi.

$$2x^{2}+3x=1-17$$

1 dan 17 ni ayirish.

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

Ikki tarafini 2 ga bo‘ling.

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

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

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

-16 ni 2 ga bo'lish.

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

\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.

$$x^{2}+\frac{3}{2}x+\left(\frac{3}{4}\right)^{2}=-8+\left(\frac{3}{4}\right)^{2}$$

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

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

-8 ni \frac{9}{16} ga qo'shish.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=-\frac{119}{16}$$

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.

$$\left(x+\frac{3}{4}\right)^{2}=-\frac{119}{16}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{3}{4}\right)^{2}}=\sqrt{-\frac{119}{16}}$$

Qisqartirish.

$$x+\frac{3}{4}=\frac{\sqrt{119}i}{4} x+\frac{3}{4}=-\frac{\sqrt{119}i}{4}$$

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

$$x=\frac{-3+\sqrt{119}i}{4} x=\frac{-\sqrt{119}i-3}{4}$$