Solve sqrt{[(1-t)-(1)]^2+[(1)-(2t)]^2}=sqrt{11-9t}

$$\sqrt { [ ( 1 - t ) - ( 1 ) ] ^ { 2 } + [ ( 1 ) - ( 2 t ) ] ^ { 2 } } = \sqrt { 11 - 9 t }$$

Solve for t

$$t=1$$

$$t=-2$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{\left(1-t-1\right)^{2}+\left(1-2t\right)^{2}}\right)^{2}=\left(\sqrt{11-9t}\right)^{2}$$

Subtract 1 from 1 to get 0.

$$\left(\sqrt{\left(-t\right)^{2}+\left(1-2t\right)^{2}}\right)^{2}=\left(\sqrt{11-9t}\right)^{2}$$

Expand \left(-t\right)^{2}.

$$\left(\sqrt{\left(-1\right)^{2}t^{2}+\left(1-2t\right)^{2}}\right)^{2}=\left(\sqrt{11-9t}\right)^{2}$$

Calculate -1 to the power of 2 and get 1.

$$\left(\sqrt{1t^{2}+\left(1-2t\right)^{2}}\right)^{2}=\left(\sqrt{11-9t}\right)^{2}$$

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

$$\left(\sqrt{1t^{2}+1-4t+4t^{2}}\right)^{2}=\left(\sqrt{11-9t}\right)^{2}$$

Combine 1t^{2} and 4t^{2} to get 5t^{2}.

$$\left(\sqrt{5t^{2}+1-4t}\right)^{2}=\left(\sqrt{11-9t}\right)^{2}$$

Calculate \sqrt{5t^{2}+1-4t} to the power of 2 and get 5t^{2}+1-4t.

$$5t^{2}+1-4t=\left(\sqrt{11-9t}\right)^{2}$$

Calculate \sqrt{11-9t} to the power of 2 and get 11-9t.

$$5t^{2}+1-4t=11-9t$$

Subtract 11 from both sides.

$$5t^{2}+1-4t-11=-9t$$

Subtract 11 from 1 to get -10.

$$5t^{2}-10-4t=-9t$$

Add 9t to both sides.

$$5t^{2}-10-4t+9t=0$$

Combine -4t and 9t to get 5t.

$$5t^{2}-10+5t=0$$

Divide both sides by 5.

$$t^{2}-2+t=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$t^{2}+t-2=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as t^{2}+at+bt-2. To find a and b, set up a system to be solved.

$$a+b=1 ab=1\left(-2\right)=-2$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=-1 b=2$$

Rewrite t^{2}+t-2 as \left(t^{2}-t\right)+\left(2t-2\right).

$$\left(t^{2}-t\right)+\left(2t-2\right)$$

Factor out t in the first and 2 in the second group.

$$t\left(t-1\right)+2\left(t-1\right)$$

Factor out common term t-1 by using distributive property.

$$\left(t-1\right)\left(t+2\right)$$

To find equation solutions, solve t-1=0 and t+2=0.

$$t=1 t=-2$$

Substitute 1 for t in the equation \sqrt{\left(1-t-1\right)^{2}+\left(1-2t\right)^{2}}=\sqrt{11-9t}.

$$\sqrt{\left(1-1-1\right)^{2}+\left(1-2\right)^{2}}=\sqrt{11-9}$$

Simplify. The value t=1 satisfies the equation.

$$2^{\frac{1}{2}}=2^{\frac{1}{2}}$$

Substitute -2 for t in the equation \sqrt{\left(1-t-1\right)^{2}+\left(1-2t\right)^{2}}=\sqrt{11-9t}.

$$\sqrt{\left(1-\left(-2\right)-1\right)^{2}+\left(1-2\left(-2\right)\right)^{2}}=\sqrt{11-9\left(-2\right)}$$

Simplify. The value t=-2 satisfies the equation.

$$29^{\frac{1}{2}}=29^{\frac{1}{2}}$$

List all solutions of \sqrt{\left(1-2t\right)^{2}+\left(-t\right)^{2}}=\sqrt{11-9t}.

$$t=1 t=-2$$