x^2-x+5=(x-a)^2+b eching

$$x ^ { 2 } - x + 5 = ( x - a ) ^ { 2 } + b$$

b uchun yechish

$$b=a\left(2x-a\right)-x+5$$

Yechim qadamlari

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

$$x^{2}-x+5=x^{2}-2xa+a^{2}+b$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$x^{2}-2xa+a^{2}+b=x^{2}-x+5$$

Ikkala tarafdan x^{2} ni ayirish.

$$-2xa+a^{2}+b=x^{2}-x+5-x^{2}$$

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

$$-2xa+a^{2}+b=-x+5$$

2xa ni ikki tarafga qo’shing.

$$a^{2}+b=-x+5+2xa$$

Ikkala tarafdan a^{2} ni ayirish.

$$b=-x+5+2xa-a^{2}$$

a uchun yechish (complex solution)

$$a=\sqrt{x^{2}-x-b+5}+x$$

$$a=-\sqrt{x^{2}-x-b+5}+x$$

a uchun yechish

$$a=\sqrt{x^{2}-x-b+5}+x$$

$$a=-\sqrt{x^{2}-x-b+5}+x\text{, }b\leq x^{2}-x+5$$