Whakaoti i te left(1-xright)^2+1/4=x^2

$${ \left(1-x \right) }^{ 2 } + \frac{ 1 }{ 4 } = { x }^{ 2 }$$

Whakaoti mō x

$$x=\frac{5}{8}=0.625$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(1-x\right)^{2}.

$$1-2x+x^{2}+\frac{1}{4}=x^{2}$$

Tāpirihia te 1 ki te \frac{1}{4}, ka \frac{5}{4}.

$$\frac{5}{4}-2x+x^{2}=x^{2}$$

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

$$\frac{5}{4}-2x+x^{2}-x^{2}=0$$

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

$$\frac{5}{4}-2x=0$$

Tangohia te \frac{5}{4} mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$-2x=-\frac{5}{4}$$

Whakawehea ngā taha e rua ki te -2.

$$x=\frac{-\frac{5}{4}}{-2}$$

Tuhia te \frac{-\frac{5}{4}}{-2} hei hautanga kotahi.

$$x=\frac{-5}{4\left(-2\right)}$$

Whakareatia te 4 ki te -2, ka -8.

$$x=\frac{-5}{-8}$$

Ka taea te hautanga \frac{-5}{-8} te whakamāmā ki te \frac{5}{8} mā te tango tahi i te tohu tōraro i te taurunga me te tauraro.

$$x=\frac{5}{8}$$