5^2+x^2=6^2 eching

$${ 5 }^{ 2 } + { x }^{ 2 } = { 6 }^{ 2 }$$

x uchun yechish

$$x=\sqrt{11}\approx 3,31662479$$

$$x=-\sqrt{11}\approx -3,31662479$$

Kvadrat ildizini topish bo'yicha qadamlar

2 daraja ko‘rsatkichini 5 ga hisoblang va 25 ni qiymatni oling.

$$25+x^{2}=6^{2}$$

2 daraja ko‘rsatkichini 6 ga hisoblang va 36 ni qiymatni oling.

$$25+x^{2}=36$$

Ikkala tarafdan 25 ni ayirish.

$$x^{2}=36-25$$

11 olish uchun 36 dan 25 ni ayirish.

$$x^{2}=11$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$x=\sqrt{11} x=-\sqrt{11}$$

Kvadrat tenglamalaridan foydalanish qoidalari

2 daraja ko‘rsatkichini 5 ga hisoblang va 25 ni qiymatni oling.

$$25+x^{2}=6^{2}$$

2 daraja ko‘rsatkichini 6 ga hisoblang va 36 ni qiymatni oling.

$$25+x^{2}=36$$

Ikkala tarafdan 36 ni ayirish.

$$25+x^{2}-36=0$$

-11 olish uchun 25 dan 36 ni ayirish.

$$-11+x^{2}=0$$

Bu kabi kvadrat tenglamalarni x^{2} sharti bilan, biroq x shartisiz hamon kvadrat tenglamasidan foydalanib yechish mumkin, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, ular standart formulaga solingandan so'ng: ax^{2}+bx+c=0.

$$x^{2}-11=0$$

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 -11 ni c bilan almashtiring.

$$x=\frac{0±\sqrt{0^{2}-4\left(-11\right)}}{2}$$

0 kvadratini chiqarish.

$$x=\frac{0±\sqrt{-4\left(-11\right)}}{2}$$

-4 ni -11 marotabaga ko'paytirish.

$$x=\frac{0±\sqrt{44}}{2}$$

44 ning kvadrat ildizini chiqarish.

$$x=\frac{0±2\sqrt{11}}{2}$$

x=\frac{0±2\sqrt{11}}{2} tenglamasini yeching, bunda ± musbat.

$$x=\sqrt{11}$$

x=\frac{0±2\sqrt{11}}{2} tenglamasini yeching, bunda ± manfiy.

$$x=-\sqrt{11}$$

Tenglama yechildi.

$$x=\sqrt{11} x=-\sqrt{11}$$