x^2+25=57 eching

$${ x }^{ 2 } +25=57$$

x uchun yechish

$$x=4\sqrt{2}\approx 5,656854249$$

$$x=-4\sqrt{2}\approx -5,656854249$$

Kvadrat ildizini topish bo'yicha qadamlar

Ikkala tarafdan 25 ni ayirish.

$$x^{2}=57-25$$

32 olish uchun 57 dan 25 ni ayirish.

$$x^{2}=32$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$x=4\sqrt{2} x=-4\sqrt{2}$$

Kvadrat tenglamalaridan foydalanish qoidalari

Ikkala tarafdan 57 ni ayirish.

$$x^{2}+25-57=0$$

-32 olish uchun 25 dan 57 ni ayirish.

$$x^{2}-32=0$$

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 -32 ni c bilan almashtiring.

$$x=\frac{0±\sqrt{0^{2}-4\left(-32\right)}}{2}$$

0 kvadratini chiqarish.

$$x=\frac{0±\sqrt{-4\left(-32\right)}}{2}$$

-4 ni -32 marotabaga ko'paytirish.

$$x=\frac{0±\sqrt{128}}{2}$$

128 ning kvadrat ildizini chiqarish.

$$x=\frac{0±8\sqrt{2}}{2}$$

x=\frac{0±8\sqrt{2}}{2} tenglamasini yeching, bunda ± musbat.

$$x=4\sqrt{2}$$

x=\frac{0±8\sqrt{2}}{2} tenglamasini yeching, bunda ± manfiy.

$$x=-4\sqrt{2}$$

Tenglama yechildi.

$$x=4\sqrt{2} x=-4\sqrt{2}$$