x^2-5x+625=8 eching

$${ x }^{ 2 } -5x+625 = 8$$

x uchun yechish (complex solution)

$$x=\frac{5+\sqrt{2443}i}{2}\approx 2,5+24,713356713i$$

$$x=\frac{-\sqrt{2443}i+5}{2}\approx 2,5-24,713356713i$$

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.

$$x^{2}-5x+625=8$$

Tenglamaning ikkala tarafidan 8 ni ayirish.

$$x^{2}-5x+625-8=8-8$$

O‘zidan 8 ayirilsa 0 qoladi.

$$x^{2}-5x+625-8=0$$

625 dan 8 ni ayirish.

$$x^{2}-5x+617=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, -5 ni b va 617 ni c bilan almashtiring.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 617}}{2}$$

-5 kvadratini chiqarish.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 617}}{2}$$

-4 ni 617 marotabaga ko'paytirish.

$$x=\frac{-\left(-5\right)±\sqrt{25-2468}}{2}$$

25 ni -2468 ga qo'shish.

$$x=\frac{-\left(-5\right)±\sqrt{-2443}}{2}$$

-2443 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-5\right)±\sqrt{2443}i}{2}$$

-5 ning teskarisi 5 ga teng.

$$x=\frac{5±\sqrt{2443}i}{2}$$

x=\frac{5±\sqrt{2443}i}{2} tenglamasini yeching, bunda ± musbat. 5 ni i\sqrt{2443} ga qo'shish.

$$x=\frac{5+\sqrt{2443}i}{2}$$

x=\frac{5±\sqrt{2443}i}{2} tenglamasini yeching, bunda ± manfiy. 5 dan i\sqrt{2443} ni ayirish.

$$x=\frac{-\sqrt{2443}i+5}{2}$$

Tenglama yechildi.

$$x=\frac{5+\sqrt{2443}i}{2} x=\frac{-\sqrt{2443}i+5}{2}$$

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.

$$x^{2}-5x+625=8$$

Tenglamaning ikkala tarafidan 625 ni ayirish.

$$x^{2}-5x+625-625=8-625$$

O‘zidan 625 ayirilsa 0 qoladi.

$$x^{2}-5x=8-625$$

8 dan 625 ni ayirish.

$$x^{2}-5x=-617$$

-5 ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{5}{2} olish uchun. Keyin, -\frac{5}{2} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=-617+\left(-\frac{5}{2}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{5}{2} kvadratini chiqarish.

$$x^{2}-5x+\frac{25}{4}=-617+\frac{25}{4}$$

-617 ni \frac{25}{4} ga qo'shish.

$$x^{2}-5x+\frac{25}{4}=-\frac{2443}{4}$$

x^{2}-5x+\frac{25}{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.

$$\left(x-\frac{5}{2}\right)^{2}=-\frac{2443}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{-\frac{2443}{4}}$$

Qisqartirish.

$$x-\frac{5}{2}=\frac{\sqrt{2443}i}{2} x-\frac{5}{2}=-\frac{\sqrt{2443}i}{2}$$

\frac{5}{2} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{5+\sqrt{2443}i}{2} x=\frac{-\sqrt{2443}i+5}{2}$$