0=x^2-4x+4+5 eching

$$0 = x ^ { 2 } - 4 x + 4 + 5$$

x uchun yechish (complex solution)

$$x=2+\sqrt{5}i\approx 2+2,236067977i$$

$$x=-\sqrt{5}i+2\approx 2-2,236067977i$$

Kvadrat tenglamalaridan foydalanish qoidalari

9 olish uchun 4 va 5'ni qo'shing.

$$0=x^{2}-4x+9$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$x^{2}-4x+9=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, -4 ni b va 9 ni c bilan almashtiring.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 9}}{2}$$

-4 kvadratini chiqarish.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 9}}{2}$$

-4 ni 9 marotabaga ko'paytirish.

$$x=\frac{-\left(-4\right)±\sqrt{16-36}}{2}$$

16 ni -36 ga qo'shish.

$$x=\frac{-\left(-4\right)±\sqrt{-20}}{2}$$

-20 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-4\right)±2\sqrt{5}i}{2}$$

-4 ning teskarisi 4 ga teng.

$$x=\frac{4±2\sqrt{5}i}{2}$$

x=\frac{4±2\sqrt{5}i}{2} tenglamasini yeching, bunda ± musbat. 4 ni 2i\sqrt{5} ga qo'shish.

$$x=\frac{4+2\sqrt{5}i}{2}$$

4+2i\sqrt{5} ni 2 ga bo'lish.

$$x=2+\sqrt{5}i$$

x=\frac{4±2\sqrt{5}i}{2} tenglamasini yeching, bunda ± manfiy. 4 dan 2i\sqrt{5} ni ayirish.

$$x=\frac{-2\sqrt{5}i+4}{2}$$

4-2i\sqrt{5} ni 2 ga bo'lish.

$$x=-\sqrt{5}i+2$$

Tenglama yechildi.

$$x=2+\sqrt{5}i x=-\sqrt{5}i+2$$

Kvadratni yakunlash bo'yicha qadamlar

9 olish uchun 4 va 5'ni qo'shing.

$$0=x^{2}-4x+9$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$x^{2}-4x+9=0$$

Ikkala tarafdan 9 ni ayirish. Har qanday sonni noldan ayirsangiz, o‘zining manfiyi chiqadi.

$$x^{2}-4x=-9$$

-4 ni bo‘lish, x shartining koeffitsienti, 2 ga -2 olish uchun. Keyin, -2 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-4x+\left(-2\right)^{2}=-9+\left(-2\right)^{2}$$

-2 kvadratini chiqarish.

$$x^{2}-4x+4=-9+4$$

-9 ni 4 ga qo'shish.

$$x^{2}-4x+4=-5$$

x^{2}-4x+4 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

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

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

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

Qisqartirish.

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

2 ni tenglamaning ikkala tarafiga qo'shish.

$$x=2+\sqrt{5}i x=-\sqrt{5}i+2$$