2x^2+8=7 eching

$$2 x ^ { 2 } + 8 = 7$$

x uchun yechish (complex solution)

$$x=-\frac{\sqrt{2}i}{2}\approx -0-0,707106781i$$

$$x=\frac{\sqrt{2}i}{2}\approx 0,707106781i$$

Kvadrat ildizini topish bo'yicha qadamlar

Ikkala tarafdan 8 ni ayirish.

$$2x^{2}=7-8$$

-1 olish uchun 7 dan 8 ni ayirish.

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

Ikki tarafini 2 ga bo‘ling.

$$x^{2}=-\frac{1}{2}$$

Tenglama yechildi.

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

Kvadrat tenglamalaridan foydalanish qoidalari

Ikkala tarafdan 7 ni ayirish.

$$2x^{2}+8-7=0$$

1 olish uchun 8 dan 7 ni ayirish.

$$2x^{2}+1=0$$

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

$$x=\frac{0±\sqrt{0^{2}-4\times 2}}{2\times 2}$$

0 kvadratini chiqarish.

$$x=\frac{0±\sqrt{-4\times 2}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{0±\sqrt{-8}}{2\times 2}$$

-8 ning kvadrat ildizini chiqarish.

$$x=\frac{0±2\sqrt{2}i}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{0±2\sqrt{2}i}{4}$$

x=\frac{0±2\sqrt{2}i}{4} tenglamasini yeching, bunda ± musbat.

$$x=\frac{\sqrt{2}i}{2}$$

x=\frac{0±2\sqrt{2}i}{4} tenglamasini yeching, bunda ± manfiy.

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

Tenglama yechildi.

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