0=1/5left(x+5right)^2-1 eching

$$0= \frac{ 1 }{ 5 } { \left(x+5 \right) }^{ 2 } -1$$

x uchun yechish

$$x=\sqrt{5}-5\approx -2,763932023$$

$$x=-\sqrt{5}-5\approx -7,236067977$$

Kvadrat tenglamalaridan foydalanish qoidalari

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(x+5\right)^{2} kengaytirilishi uchun ishlating.

$$0=\frac{1}{5}\left(x^{2}+10x+25\right)-1$$

\frac{1}{5} ga x^{2}+10x+25 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$0=\frac{1}{5}x^{2}+2x+5-1$$

4 olish uchun 5 dan 1 ni ayirish.

$$0=\frac{1}{5}x^{2}+2x+4$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$\frac{1}{5}x^{2}+2x+4=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} \frac{1}{5} ni a, 2 ni b va 4 ni c bilan almashtiring.

$$x=\frac{-2±\sqrt{2^{2}-4\times \frac{1}{5}\times 4}}{2\times \frac{1}{5}}$$

2 kvadratini chiqarish.

$$x=\frac{-2±\sqrt{4-4\times \frac{1}{5}\times 4}}{2\times \frac{1}{5}}$$

-4 ni \frac{1}{5} marotabaga ko'paytirish.

$$x=\frac{-2±\sqrt{4-\frac{4}{5}\times 4}}{2\times \frac{1}{5}}$$

-\frac{4}{5} ni 4 marotabaga ko'paytirish.

$$x=\frac{-2±\sqrt{4-\frac{16}{5}}}{2\times \frac{1}{5}}$$

4 ni -\frac{16}{5} ga qo'shish.

$$x=\frac{-2±\sqrt{\frac{4}{5}}}{2\times \frac{1}{5}}$$

\frac{4}{5} ning kvadrat ildizini chiqarish.

$$x=\frac{-2±\frac{2\sqrt{5}}{5}}{2\times \frac{1}{5}}$$

2 ni \frac{1}{5} marotabaga ko'paytirish.

$$x=\frac{-2±\frac{2\sqrt{5}}{5}}{\frac{2}{5}}$$

x=\frac{-2±\frac{2\sqrt{5}}{5}}{\frac{2}{5}} tenglamasini yeching, bunda ± musbat. -2 ni \frac{2\sqrt{5}}{5} ga qo'shish.

$$x=\frac{\frac{2\sqrt{5}}{5}-2}{\frac{2}{5}}$$

-2+\frac{2\sqrt{5}}{5} ni \frac{2}{5} ga bo'lish -2+\frac{2\sqrt{5}}{5} ga k'paytirish \frac{2}{5} ga qaytarish.

$$x=\sqrt{5}-5$$

x=\frac{-2±\frac{2\sqrt{5}}{5}}{\frac{2}{5}} tenglamasini yeching, bunda ± manfiy. -2 dan \frac{2\sqrt{5}}{5} ni ayirish.

$$x=\frac{-\frac{2\sqrt{5}}{5}-2}{\frac{2}{5}}$$

-2-\frac{2\sqrt{5}}{5} ni \frac{2}{5} ga bo'lish -2-\frac{2\sqrt{5}}{5} ga k'paytirish \frac{2}{5} ga qaytarish.

$$x=-\sqrt{5}-5$$

Tenglama yechildi.

$$x=\sqrt{5}-5 x=-\sqrt{5}-5$$

Kvadratni yakunlash bo'yicha qadamlar

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(x+5\right)^{2} kengaytirilishi uchun ishlating.

$$0=\frac{1}{5}\left(x^{2}+10x+25\right)-1$$

\frac{1}{5} ga x^{2}+10x+25 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$0=\frac{1}{5}x^{2}+2x+5-1$$

4 olish uchun 5 dan 1 ni ayirish.

$$0=\frac{1}{5}x^{2}+2x+4$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$\frac{1}{5}x^{2}+2x+4=0$$

Ikkala tarafdan 4 ni ayirish. Har qanday sonni noldan ayirsangiz, o‘zining manfiyi chiqadi.

$$\frac{1}{5}x^{2}+2x=-4$$

Ikkala tarafini 5 ga ko‘paytiring.

$$\frac{\frac{1}{5}x^{2}+2x}{\frac{1}{5}}=-\frac{4}{\frac{1}{5}}$$

\frac{1}{5} ga bo'lish \frac{1}{5} ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{2}{\frac{1}{5}}x=-\frac{4}{\frac{1}{5}}$$

2 ni \frac{1}{5} ga bo'lish 2 ga k'paytirish \frac{1}{5} ga qaytarish.

$$x^{2}+10x=-\frac{4}{\frac{1}{5}}$$

-4 ni \frac{1}{5} ga bo'lish -4 ga k'paytirish \frac{1}{5} ga qaytarish.

$$x^{2}+10x=-20$$

10 ni bo‘lish, x shartining koeffitsienti, 2 ga 5 olish uchun. Keyin, 5 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+10x+5^{2}=-20+5^{2}$$

5 kvadratini chiqarish.

$$x^{2}+10x+25=-20+25$$

-20 ni 25 ga qo'shish.

$$x^{2}+10x+25=5$$

x^{2}+10x+25 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+5\right)^{2}=5$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+5\right)^{2}}=\sqrt{5}$$

Qisqartirish.

$$x+5=\sqrt{5} x+5=-\sqrt{5}$$

Tenglamaning ikkala tarafidan 5 ni ayirish.

$$x=\sqrt{5}-5 x=-\sqrt{5}-5$$