x^2+3x+4=(x+a)^2+b eching

$$x ^ { 2 } + 3 x + 4 = ( x + a ) ^ { 2 } + b$$

b uchun yechish

$$b=-a\left(2x+a\right)+3x+4$$

Yechim qadamlari

\left(p+q\right)^{2}=p^{2}+2pq+q^{2} binom teoremasini \left(x+a\right)^{2} kengaytirilishi uchun ishlating.

$$x^{2}+3x+4=x^{2}+2xa+a^{2}+b$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$x^{2}+2xa+a^{2}+b=x^{2}+3x+4$$

Ikkala tarafdan x^{2} ni ayirish.

$$2xa+a^{2}+b=x^{2}+3x+4-x^{2}$$

0 ni olish uchun x^{2} va -x^{2} ni birlashtirish.

$$2xa+a^{2}+b=3x+4$$

Ikkala tarafdan 2xa ni ayirish.

$$a^{2}+b=3x+4-2xa$$

Ikkala tarafdan a^{2} ni ayirish.

$$b=3x+4-2xa-a^{2}$$

a uchun yechish (complex solution)

$$a=-\left(\sqrt{x^{2}+3x-b+4}+x\right)$$

$$a=\sqrt{x^{2}+3x-b+4}-x$$

a uchun yechish

$$a=-\left(\sqrt{x^{2}+3x-b+4}+x\right)$$

$$a=\sqrt{x^{2}+3x-b+4}-x\text{, }b\leq x^{2}+3x+4$$