3^2=25^2+x^2 eching

$${ 3 }^{ 2 } = { 25 }^{ 2 } + { x }^{ 2 }$$

x uchun yechish (complex solution)

$$x=-2\sqrt{154}i\approx -0-24,819347292i$$

$$x=2\sqrt{154}i\approx 24,819347292i$$

Kvadrat ildizini topish bo'yicha qadamlar

2 daraja ko‘rsatkichini 3 ga hisoblang va 9 ni qiymatni oling.

$$9=25^{2}+x^{2}$$

2 daraja ko‘rsatkichini 25 ga hisoblang va 625 ni qiymatni oling.

$$9=625+x^{2}$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$625+x^{2}=9$$

Ikkala tarafdan 625 ni ayirish.

$$x^{2}=9-625$$

-616 olish uchun 9 dan 625 ni ayirish.

$$x^{2}=-616$$

Tenglama yechildi.

$$x=2\sqrt{154}i x=-2\sqrt{154}i$$

Kvadrat tenglamalaridan foydalanish qoidalari

2 daraja ko‘rsatkichini 3 ga hisoblang va 9 ni qiymatni oling.

$$9=25^{2}+x^{2}$$

2 daraja ko‘rsatkichini 25 ga hisoblang va 625 ni qiymatni oling.

$$9=625+x^{2}$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$625+x^{2}=9$$

Ikkala tarafdan 9 ni ayirish.

$$625+x^{2}-9=0$$

616 olish uchun 625 dan 9 ni ayirish.

$$616+x^{2}=0$$

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.

$$x^{2}+616=0$$

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 616 ni c bilan almashtiring.

$$x=\frac{0±\sqrt{0^{2}-4\times 616}}{2}$$

0 kvadratini chiqarish.

$$x=\frac{0±\sqrt{-4\times 616}}{2}$$

-4 ni 616 marotabaga ko'paytirish.

$$x=\frac{0±\sqrt{-2464}}{2}$$

-2464 ning kvadrat ildizini chiqarish.

$$x=\frac{0±4\sqrt{154}i}{2}$$

x=\frac{0±4\sqrt{154}i}{2} tenglamasini yeching, bunda ± musbat.

$$x=2\sqrt{154}i$$

x=\frac{0±4\sqrt{154}i}{2} tenglamasini yeching, bunda ± manfiy.

$$x=-2\sqrt{154}i$$

Tenglama yechildi.

$$x=2\sqrt{154}i x=-2\sqrt{154}i$$