x^2-5x+16=0 eching

$$x ^ { 2 } - 5 x + 16 = 0$$

x uchun yechish (complex solution)

$$x=\frac{5+\sqrt{39}i}{2}\approx 2,5+3,122498999i$$

$$x=\frac{-\sqrt{39}i+5}{2}\approx 2,5-3,122498999i$$

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.

$$x^{2}-5x+16=0$$

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 16 ni c bilan almashtiring.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 16}}{2}$$

-5 kvadratini chiqarish.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 16}}{2}$$

-4 ni 16 marotabaga ko'paytirish.

$$x=\frac{-\left(-5\right)±\sqrt{25-64}}{2}$$

25 ni -64 ga qo'shish.

$$x=\frac{-\left(-5\right)±\sqrt{-39}}{2}$$

-39 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-5\right)±\sqrt{39}i}{2}$$

-5 ning teskarisi 5 ga teng.

$$x=\frac{5±\sqrt{39}i}{2}$$

x=\frac{5±\sqrt{39}i}{2} tenglamasini yeching, bunda ± musbat. 5 ni i\sqrt{39} ga qo'shish.

$$x=\frac{5+\sqrt{39}i}{2}$$

x=\frac{5±\sqrt{39}i}{2} tenglamasini yeching, bunda ± manfiy. 5 dan i\sqrt{39} ni ayirish.

$$x=\frac{-\sqrt{39}i+5}{2}$$

Tenglama yechildi.

$$x=\frac{5+\sqrt{39}i}{2} x=\frac{-\sqrt{39}i+5}{2}$$

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.

$$x^{2}-5x+16=0$$

Tenglamaning ikkala tarafidan 16 ni ayirish.

$$x^{2}-5x+16-16=-16$$

O‘zidan 16 ayirilsa 0 qoladi.

$$x^{2}-5x=-16$$

-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.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=-16+\left(-\frac{5}{2}\right)^{2}$$

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

$$x^{2}-5x+\frac{25}{4}=-16+\frac{25}{4}$$

-16 ni \frac{25}{4} ga qo'shish.

$$x^{2}-5x+\frac{25}{4}=-\frac{39}{4}$$

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.

$$\left(x-\frac{5}{2}\right)^{2}=-\frac{39}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{-\frac{39}{4}}$$

Qisqartirish.

$$x-\frac{5}{2}=\frac{\sqrt{39}i}{2} x-\frac{5}{2}=-\frac{\sqrt{39}i}{2}$$

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

$$x=\frac{5+\sqrt{39}i}{2} x=\frac{-\sqrt{39}i+5}{2}$$