(12-5)^2+(7-6)^2=r^2 eching

$$( 12 - 5 ) ^ { 2 } + ( 7 - 6 ) ^ { 2 } = r ^ { 2 }$$

r uchun yechish

$$r=5\sqrt{2}\approx 7,071067812$$

$$r=-5\sqrt{2}\approx -7,071067812$$

Kvadrat ildizini topish bo'yicha qadamlar

7 olish uchun 12 dan 5 ni ayirish.

$$7^{2}+\left(7-6\right)^{2}=r^{2}$$

2 daraja ko‘rsatkichini 7 ga hisoblang va 49 ni qiymatni oling.

$$49+\left(7-6\right)^{2}=r^{2}$$

1 olish uchun 7 dan 6 ni ayirish.

$$49+1^{2}=r^{2}$$

2 daraja ko‘rsatkichini 1 ga hisoblang va 1 ni qiymatni oling.

$$49+1=r^{2}$$

50 olish uchun 49 va 1'ni qo'shing.

$$50=r^{2}$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$r^{2}=50$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$r=5\sqrt{2} r=-5\sqrt{2}$$

Kvadrat tenglamalaridan foydalanish qoidalari

7 olish uchun 12 dan 5 ni ayirish.

$$7^{2}+\left(7-6\right)^{2}=r^{2}$$

2 daraja ko‘rsatkichini 7 ga hisoblang va 49 ni qiymatni oling.

$$49+\left(7-6\right)^{2}=r^{2}$$

1 olish uchun 7 dan 6 ni ayirish.

$$49+1^{2}=r^{2}$$

2 daraja ko‘rsatkichini 1 ga hisoblang va 1 ni qiymatni oling.

$$49+1=r^{2}$$

50 olish uchun 49 va 1'ni qo'shing.

$$50=r^{2}$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$r^{2}=50$$

Ikkala tarafdan 50 ni ayirish.

$$r^{2}-50=0$$

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 -50 ni c bilan almashtiring.

$$r=\frac{0±\sqrt{0^{2}-4\left(-50\right)}}{2}$$

0 kvadratini chiqarish.

$$r=\frac{0±\sqrt{-4\left(-50\right)}}{2}$$

-4 ni -50 marotabaga ko'paytirish.

$$r=\frac{0±\sqrt{200}}{2}$$

200 ning kvadrat ildizini chiqarish.

$$r=\frac{0±10\sqrt{2}}{2}$$

r=\frac{0±10\sqrt{2}}{2} tenglamasini yeching, bunda ± musbat.

$$r=5\sqrt{2}$$

r=\frac{0±10\sqrt{2}}{2} tenglamasini yeching, bunda ± manfiy.

$$r=-5\sqrt{2}$$

Tenglama yechildi.

$$r=5\sqrt{2} r=-5\sqrt{2}$$