z^2+16z+64=7 eching

$$z ^ { 2 } + 16 z + 64 = 7$$

z uchun yechish

$$z=\sqrt{7}-8\approx -5,354248689$$

$$z=-\sqrt{7}-8\approx -10,645751311$$

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.

$$z^{2}+16z+64=7$$

Tenglamaning ikkala tarafidan 7 ni ayirish.

$$z^{2}+16z+64-7=7-7$$

O‘zidan 7 ayirilsa 0 qoladi.

$$z^{2}+16z+64-7=0$$

64 dan 7 ni ayirish.

$$z^{2}+16z+57=0$$

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 57 ni c bilan almashtiring.

$$z=\frac{-16±\sqrt{16^{2}-4\times 57}}{2}$$

16 kvadratini chiqarish.

$$z=\frac{-16±\sqrt{256-4\times 57}}{2}$$

-4 ni 57 marotabaga ko'paytirish.

$$z=\frac{-16±\sqrt{256-228}}{2}$$

256 ni -228 ga qo'shish.

$$z=\frac{-16±\sqrt{28}}{2}$$

28 ning kvadrat ildizini chiqarish.

$$z=\frac{-16±2\sqrt{7}}{2}$$

z=\frac{-16±2\sqrt{7}}{2} tenglamasini yeching, bunda ± musbat. -16 ni 2\sqrt{7} ga qo'shish.

$$z=\frac{2\sqrt{7}-16}{2}$$

-16+2\sqrt{7} ni 2 ga bo'lish.

$$z=\sqrt{7}-8$$

z=\frac{-16±2\sqrt{7}}{2} tenglamasini yeching, bunda ± manfiy. -16 dan 2\sqrt{7} ni ayirish.

$$z=\frac{-2\sqrt{7}-16}{2}$$

-16-2\sqrt{7} ni 2 ga bo'lish.

$$z=-\sqrt{7}-8$$

Tenglama yechildi.

$$z=\sqrt{7}-8 z=-\sqrt{7}-8$$

Kvadratni yakunlash bo'yicha qadamlar

z^{2}+16z+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.

$$\left(z+8\right)^{2}=7$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(z+8\right)^{2}}=\sqrt{7}$$

Qisqartirish.

$$z+8=\sqrt{7} z+8=-\sqrt{7}$$

Tenglamaning ikkala tarafidan 8 ni ayirish.

$$z=\sqrt{7}-8 z=-\sqrt{7}-8$$