Solve t^2=4(1-t)^2

$$t ^ { 2 } = 4 ( 1 - t ) ^ { 2 }$$

Solve for t

$$t=2$$

$$t=\frac{2}{3}\approx 0.666666667$$

Steps Using Factoring By Grouping

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

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

Use the distributive property to multiply 4 by 1-2t+t^{2}.

$$t^{2}=4-8t+4t^{2}$$

Subtract 4 from both sides.

$$t^{2}-4=-8t+4t^{2}$$

Add 8t to both sides.

$$t^{2}-4+8t=4t^{2}$$

Subtract 4t^{2} from both sides.

$$t^{2}-4+8t-4t^{2}=0$$

Combine t^{2} and -4t^{2} to get -3t^{2}.

$$-3t^{2}-4+8t=0$$

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

$$-3t^{2}+8t-4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -3t^{2}+at+bt-4. To find a and b, set up a system to be solved.

$$a+b=8 ab=-3\left(-4\right)=12$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 12.

$$1,12 2,6 3,4$$

Calculate the sum for each pair.

$$1+12=13 2+6=8 3+4=7$$

The solution is the pair that gives sum 8.

$$a=6 b=2$$

Rewrite -3t^{2}+8t-4 as \left(-3t^{2}+6t\right)+\left(2t-4\right).

$$\left(-3t^{2}+6t\right)+\left(2t-4\right)$$

Factor out 3t in the first and -2 in the second group.

$$3t\left(-t+2\right)-2\left(-t+2\right)$$

Factor out common term -t+2 by using distributive property.

$$\left(-t+2\right)\left(3t-2\right)$$

To find equation solutions, solve -t+2=0 and 3t-2=0.

$$t=2 t=\frac{2}{3}$$

Steps Using the Quadratic Formula

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

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

Use the distributive property to multiply 4 by 1-2t+t^{2}.

$$t^{2}=4-8t+4t^{2}$$

Subtract 4 from both sides.

$$t^{2}-4=-8t+4t^{2}$$

Add 8t to both sides.

$$t^{2}-4+8t=4t^{2}$$

Subtract 4t^{2} from both sides.

$$t^{2}-4+8t-4t^{2}=0$$

Combine t^{2} and -4t^{2} to get -3t^{2}.

$$-3t^{2}-4+8t=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-3t^{2}+8t-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 8 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-8±\sqrt{8^{2}-4\left(-3\right)\left(-4\right)}}{2\left(-3\right)}$$

Square 8.

$$t=\frac{-8±\sqrt{64-4\left(-3\right)\left(-4\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$t=\frac{-8±\sqrt{64+12\left(-4\right)}}{2\left(-3\right)}$$

Multiply 12 times -4.

$$t=\frac{-8±\sqrt{64-48}}{2\left(-3\right)}$$

Add 64 to -48.

$$t=\frac{-8±\sqrt{16}}{2\left(-3\right)}$$

Take the square root of 16.

$$t=\frac{-8±4}{2\left(-3\right)}$$

Multiply 2 times -3.

$$t=\frac{-8±4}{-6}$$

Now solve the equation t=\frac{-8±4}{-6} when ± is plus. Add -8 to 4.

$$t=-\frac{4}{-6}$$

Reduce the fraction \frac{-4}{-6} to lowest terms by extracting and canceling out 2.

$$t=\frac{2}{3}$$

Now solve the equation t=\frac{-8±4}{-6} when ± is minus. Subtract 4 from -8.

$$t=-\frac{12}{-6}$$

Divide -12 by -6.

$$t=2$$

The equation is now solved.

$$t=\frac{2}{3} t=2$$

Steps for Completing the Square

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

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

Use the distributive property to multiply 4 by 1-2t+t^{2}.

$$t^{2}=4-8t+4t^{2}$$

Add 8t to both sides.

$$t^{2}+8t=4+4t^{2}$$

Subtract 4t^{2} from both sides.

$$t^{2}+8t-4t^{2}=4$$

Combine t^{2} and -4t^{2} to get -3t^{2}.

$$-3t^{2}+8t=4$$

Divide both sides by -3.

$$\frac{-3t^{2}+8t}{-3}=\frac{4}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$t^{2}+\frac{8}{-3}t=\frac{4}{-3}$$

Divide 8 by -3.

$$t^{2}-\frac{8}{3}t=\frac{4}{-3}$$

Divide 4 by -3.

$$t^{2}-\frac{8}{3}t=-\frac{4}{3}$$

Divide -\frac{8}{3}, the coefficient of the x term, by 2 to get -\frac{4}{3}. Then add the square of -\frac{4}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-\frac{8}{3}t+\left(-\frac{4}{3}\right)^{2}=-\frac{4}{3}+\left(-\frac{4}{3}\right)^{2}$$

Square -\frac{4}{3} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-\frac{8}{3}t+\frac{16}{9}=-\frac{4}{3}+\frac{16}{9}$$

Add -\frac{4}{3} to \frac{16}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t^{2}-\frac{8}{3}t+\frac{16}{9}=\frac{4}{9}$$

Factor t^{2}-\frac{8}{3}t+\frac{16}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-\frac{4}{3}\right)^{2}=\frac{4}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{4}{3}\right)^{2}}=\sqrt{\frac{4}{9}}$$

Simplify.

$$t-\frac{4}{3}=\frac{2}{3} t-\frac{4}{3}=-\frac{2}{3}$$

Add \frac{4}{3} to both sides of the equation.

$$t=2 t=\frac{2}{3}$$