Whakaoti i te b^2-(5-x)^2=5^2-x^2

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

Whakaoti mō x

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

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(5-x\right)^{2}.

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

Hei kimi i te tauaro o 25-10x+x^{2}, kimihia te tauaro o ia taurangi.

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

Tātaihia te 5 mā te pū o 2, kia riro ko 25.

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

Me tāpiri te x^{2} ki ngā taha e rua.

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

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

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

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

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

Me tāpiri te 25 ki ngā taha e rua.

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

Tāpirihia te 25 ki te 25, ka 50.

$$10x=50-b^{2}$$

Whakawehea ngā taha e rua ki te 10.

$$\frac{10x}{10}=\frac{50-b^{2}}{10}$$

Mā te whakawehe ki te 10 ka wetekia te whakareanga ki te 10.

$$x=\frac{50-b^{2}}{10}$$

Whakawehe 50-b^{2} ki te 10.

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

Whakaoti mō b (complex solution)

$$b=-\sqrt{50-10x}$$

$$b=\sqrt{50-10x}$$

Whakaoti mō b

$$b=\sqrt{50-10x}$$

$$b=-\sqrt{50-10x}\text{, }x\leq 5$$