Solve DT=sqrt{(1-sqrt{2})^2}

$$D T = \sqrt { ( 1 - \sqrt { 2 } ) ^ { 2 } }$$

Solve for D

$$D=\frac{\sqrt{2}-1}{T}$$

$$T\neq 0$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-\sqrt{2}\right)^{2}.

$$DT=\sqrt{1-2\sqrt{2}+\left(\sqrt{2}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$DT=\sqrt{1-2\sqrt{2}+2}$$

Add 1 and 2 to get 3.

$$DT=\sqrt{3-2\sqrt{2}}$$

The equation is in standard form.

$$TD=\sqrt{3-2\sqrt{2}}$$

Divide both sides by T.

$$\frac{TD}{T}=\frac{\sqrt{2}-1}{T}$$

Dividing by T undoes the multiplication by T.

$$D=\frac{\sqrt{2}-1}{T}$$

Solve for T

$$T=\frac{\sqrt{2}-1}{D}$$

$$D\neq 0$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-\sqrt{2}\right)^{2}.

$$DT=\sqrt{1-2\sqrt{2}+\left(\sqrt{2}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$DT=\sqrt{1-2\sqrt{2}+2}$$

Add 1 and 2 to get 3.

$$DT=\sqrt{3-2\sqrt{2}}$$

Divide both sides by D.

$$\frac{DT}{D}=\frac{\sqrt{2}-1}{D}$$

Dividing by D undoes the multiplication by D.

$$T=\frac{\sqrt{2}-1}{D}$$