x^2-3x+8=1 eching

$$x ^ { 2 } - 3 x + 8 = 1$$

x uchun yechish (complex solution)

$$x=\frac{3+\sqrt{19}i}{2}\approx 1,5+2,179449472i$$

$$x=\frac{-\sqrt{19}i+3}{2}\approx 1,5-2,179449472i$$

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.

$$x^{2}-3x+8=1$$

Tenglamaning ikkala tarafidan 1 ni ayirish.

$$x^{2}-3x+8-1=1-1$$

O‘zidan 1 ayirilsa 0 qoladi.

$$x^{2}-3x+8-1=0$$

8 dan 1 ni ayirish.

$$x^{2}-3x+7=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, -3 ni b va 7 ni c bilan almashtiring.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 7}}{2}$$

-3 kvadratini chiqarish.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 7}}{2}$$

-4 ni 7 marotabaga ko'paytirish.

$$x=\frac{-\left(-3\right)±\sqrt{9-28}}{2}$$

9 ni -28 ga qo'shish.

$$x=\frac{-\left(-3\right)±\sqrt{-19}}{2}$$

-19 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-3\right)±\sqrt{19}i}{2}$$

-3 ning teskarisi 3 ga teng.

$$x=\frac{3±\sqrt{19}i}{2}$$

x=\frac{3±\sqrt{19}i}{2} tenglamasini yeching, bunda ± musbat. 3 ni i\sqrt{19} ga qo'shish.

$$x=\frac{3+\sqrt{19}i}{2}$$

x=\frac{3±\sqrt{19}i}{2} tenglamasini yeching, bunda ± manfiy. 3 dan i\sqrt{19} ni ayirish.

$$x=\frac{-\sqrt{19}i+3}{2}$$

Tenglama yechildi.

$$x=\frac{3+\sqrt{19}i}{2} x=\frac{-\sqrt{19}i+3}{2}$$

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.

$$x^{2}-3x+8=1$$

Tenglamaning ikkala tarafidan 8 ni ayirish.

$$x^{2}-3x+8-8=1-8$$

O‘zidan 8 ayirilsa 0 qoladi.

$$x^{2}-3x=1-8$$

1 dan 8 ni ayirish.

$$x^{2}-3x=-7$$

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

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=-7+\left(-\frac{3}{2}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{3}{2} kvadratini chiqarish.

$$x^{2}-3x+\frac{9}{4}=-7+\frac{9}{4}$$

-7 ni \frac{9}{4} ga qo'shish.

$$x^{2}-3x+\frac{9}{4}=-\frac{19}{4}$$

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

$$\left(x-\frac{3}{2}\right)^{2}=-\frac{19}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{-\frac{19}{4}}$$

Qisqartirish.

$$x-\frac{3}{2}=\frac{\sqrt{19}i}{2} x-\frac{3}{2}=-\frac{\sqrt{19}i}{2}$$

\frac{3}{2} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{3+\sqrt{19}i}{2} x=\frac{-\sqrt{19}i+3}{2}$$