Whakaoti i te x(x+a)+25=(x+b)^2

$$x ( x + a ) + 25 = ( x + b ) ^ { 2 }$$

Whakaoti mō a (complex solution)

$$\left\{\begin{matrix}a=\frac{2bx+b^{2}-25}{x}\text{, }&x\neq 0\\a\in \mathrm{C}\text{, }&\left(b=-5\text{ or }b=5\right)\text{ and }x=0\end{matrix}\right.$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakamahia te āhuatanga tohatoha hei whakarea te x ki te x+a.

$$x^{2}+xa+25=\left(x+b\right)^{2}$$

Whakamahia te ture huarua \left(p+q\right)^{2}=p^{2}+2pq+q^{2} hei whakaroha \left(x+b\right)^{2}.

$$x^{2}+xa+25=x^{2}+2xb+b^{2}$$

Tangohia te x^{2} mai i ngā taha e rua.

$$xa+25=x^{2}+2xb+b^{2}-x^{2}$$

Pahekotia te x^{2} me -x^{2}, ka 0.

$$xa+25=2xb+b^{2}$$

Tangohia te 25 mai i ngā taha e rua.

$$xa=2xb+b^{2}-25$$

He hanga arowhānui tō te whārite.

$$xa=2bx+b^{2}-25$$

Whakawehea ngā taha e rua ki te x.

$$\frac{xa}{x}=\frac{2bx+b^{2}-25}{x}$$

Mā te whakawehe ki te x ka wetekia te whakareanga ki te x.

$$a=\frac{2bx+b^{2}-25}{x}$$

Whakaoti mō a

$$\left\{\begin{matrix}a=\frac{2bx+b^{2}-25}{x}\text{, }&x\neq 0\\a\in \mathrm{R}\text{, }&x=0\text{ and }|b|=5\end{matrix}\right.$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakamahia te āhuatanga tohatoha hei whakarea te x ki te x+a.

$$x^{2}+xa+25=\left(x+b\right)^{2}$$

Whakamahia te ture huarua \left(p+q\right)^{2}=p^{2}+2pq+q^{2} hei whakaroha \left(x+b\right)^{2}.

$$x^{2}+xa+25=x^{2}+2xb+b^{2}$$

Tangohia te x^{2} mai i ngā taha e rua.

$$xa+25=x^{2}+2xb+b^{2}-x^{2}$$

Pahekotia te x^{2} me -x^{2}, ka 0.

$$xa+25=2xb+b^{2}$$

Tangohia te 25 mai i ngā taha e rua.

$$xa=2xb+b^{2}-25$$

He hanga arowhānui tō te whārite.

$$xa=2bx+b^{2}-25$$

Whakawehea ngā taha e rua ki te x.

$$\frac{xa}{x}=\frac{2bx+b^{2}-25}{x}$$

Mā te whakawehe ki te x ka wetekia te whakareanga ki te x.

$$a=\frac{2bx+b^{2}-25}{x}$$

Whakaoti mō b (complex solution)

$$b=-\left(\sqrt{x^{2}+ax+25}+x\right)$$

$$b=\sqrt{x^{2}+ax+25}-x$$