x^2+3-4^2=0 eching

$$x ^ { 2 } + 3 - 4 ^ { 2 } = 0$$

x uchun yechish

$$x=\sqrt{13}\approx 3,605551275$$

$$x=-\sqrt{13}\approx -3,605551275$$

Kvadrat ildizini topish bo'yicha qadamlar

2 daraja ko‘rsatkichini 4 ga hisoblang va 16 ni qiymatni oling.

$$x^{2}+3-16=0$$

-13 olish uchun 3 dan 16 ni ayirish.

$$x^{2}-13=0$$

13 ni ikki tarafga qo’shing. Har qanday songa nolni qo‘shsangiz, o‘zi chiqadi.

$$x^{2}=13$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$x=\sqrt{13} x=-\sqrt{13}$$

Kvadrat tenglamalaridan foydalanish qoidalari

2 daraja ko‘rsatkichini 4 ga hisoblang va 16 ni qiymatni oling.

$$x^{2}+3-16=0$$

-13 olish uchun 3 dan 16 ni ayirish.

$$x^{2}-13=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, 0 ni b va -13 ni c bilan almashtiring.

$$x=\frac{0±\sqrt{0^{2}-4\left(-13\right)}}{2}$$

0 kvadratini chiqarish.

$$x=\frac{0±\sqrt{-4\left(-13\right)}}{2}$$

-4 ni -13 marotabaga ko'paytirish.

$$x=\frac{0±\sqrt{52}}{2}$$

52 ning kvadrat ildizini chiqarish.

$$x=\frac{0±2\sqrt{13}}{2}$$

x=\frac{0±2\sqrt{13}}{2} tenglamasini yeching, bunda ± musbat.

$$x=\sqrt{13}$$

x=\frac{0±2\sqrt{13}}{2} tenglamasini yeching, bunda ± manfiy.

$$x=-\sqrt{13}$$

Tenglama yechildi.

$$x=\sqrt{13} x=-\sqrt{13}$$