Whakaoti i te x^2+b^2=(10-x)^2

$$x ^ { 2 } + b ^ { 2 } = ( 10 - x ) ^ { 2 }$$

Whakaoti mō x

$$x=-\frac{b^{2}}{20}+5$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(10-x\right)^{2}.

$$x^{2}+b^{2}=100-20x+x^{2}$$

Me tāpiri te 20x ki ngā taha e rua.

$$x^{2}+b^{2}+20x=100+x^{2}$$

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

$$x^{2}+b^{2}+20x-x^{2}=100$$

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

$$b^{2}+20x=100$$

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

$$20x=100-b^{2}$$

Whakawehea ngā taha e rua ki te 20.

$$\frac{20x}{20}=\frac{100-b^{2}}{20}$$

Mā te whakawehe ki te 20 ka wetekia te whakareanga ki te 20.

$$x=\frac{100-b^{2}}{20}$$

Whakawehe -b^{2}+100 ki te 20.

$$x=-\frac{b^{2}}{20}+5$$

Whakaoti mō b (complex solution)

$$b=-2\sqrt{25-5x}$$

$$b=2\sqrt{25-5x}$$

Whakaoti mō b

$$b=2\sqrt{25-5x}$$

$$b=-2\sqrt{25-5x}\text{, }x\leq 5$$