x^2-16x+54=0 eching

$$x ^ { 2 } - 16 x + 54 = 0$$

x uchun yechish

$$x=\sqrt{10}+8\approx 11,16227766$$

$$x=8-\sqrt{10}\approx 4,83772234$$

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.

$$x^{2}-16x+54=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, -16 ni b va 54 ni c bilan almashtiring.

$$x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 54}}{2}$$

-16 kvadratini chiqarish.

$$x=\frac{-\left(-16\right)±\sqrt{256-4\times 54}}{2}$$

-4 ni 54 marotabaga ko'paytirish.

$$x=\frac{-\left(-16\right)±\sqrt{256-216}}{2}$$

256 ni -216 ga qo'shish.

$$x=\frac{-\left(-16\right)±\sqrt{40}}{2}$$

40 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-16\right)±2\sqrt{10}}{2}$$

-16 ning teskarisi 16 ga teng.

$$x=\frac{16±2\sqrt{10}}{2}$$

x=\frac{16±2\sqrt{10}}{2} tenglamasini yeching, bunda ± musbat. 16 ni 2\sqrt{10} ga qo'shish.

$$x=\frac{2\sqrt{10}+16}{2}$$

16+2\sqrt{10} ni 2 ga bo'lish.

$$x=\sqrt{10}+8$$

x=\frac{16±2\sqrt{10}}{2} tenglamasini yeching, bunda ± manfiy. 16 dan 2\sqrt{10} ni ayirish.

$$x=\frac{16-2\sqrt{10}}{2}$$

16-2\sqrt{10} ni 2 ga bo'lish.

$$x=8-\sqrt{10}$$

Tenglama yechildi.

$$x=\sqrt{10}+8 x=8-\sqrt{10}$$

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.

$$x^{2}-16x+54=0$$

Tenglamaning ikkala tarafidan 54 ni ayirish.

$$x^{2}-16x+54-54=-54$$

O‘zidan 54 ayirilsa 0 qoladi.

$$x^{2}-16x=-54$$

-16 ni bo‘lish, x shartining koeffitsienti, 2 ga -8 olish uchun. Keyin, -8 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-16x+\left(-8\right)^{2}=-54+\left(-8\right)^{2}$$

-8 kvadratini chiqarish.

$$x^{2}-16x+64=-54+64$$

-54 ni 64 ga qo'shish.

$$x^{2}-16x+64=10$$

x^{2}-16x+64 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-8\right)^{2}=10$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-8\right)^{2}}=\sqrt{10}$$

Qisqartirish.

$$x-8=\sqrt{10} x-8=-\sqrt{10}$$

8 ni tenglamaning ikkala tarafiga qo'shish.

$$x=\sqrt{10}+8 x=8-\sqrt{10}$$