Solve (x+3)^2=(1-2x)^2

$$( x + 3 ) ^ { 2 } = ( 1 - 2 x ) ^ { 2 }$$

Solve for x

$$x=4$$

$$x=-\frac{2}{3}\approx -0.666666667$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$x^{2}+6x+9=\left(1-2x\right)^{2}$$

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

$$x^{2}+6x+9=1-4x+4x^{2}$$

Subtract 1 from both sides.

$$x^{2}+6x+9-1=-4x+4x^{2}$$

Subtract 1 from 9 to get 8.

$$x^{2}+6x+8=-4x+4x^{2}$$

Add 4x to both sides.

$$x^{2}+6x+8+4x=4x^{2}$$

Combine 6x and 4x to get 10x.

$$x^{2}+10x+8=4x^{2}$$

Subtract 4x^{2} from both sides.

$$x^{2}+10x+8-4x^{2}=0$$

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

$$-3x^{2}+10x+8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -3x^{2}+ax+bx+8. To find a and b, set up a system to be solved.

$$a+b=10 ab=-3\times 8=-24$$

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. List all such integer pairs that give product -24.

$$-1,24 -2,12 -3,8 -4,6$$

Calculate the sum for each pair.

$$-1+24=23 -2+12=10 -3+8=5 -4+6=2$$

The solution is the pair that gives sum 10.

$$a=12 b=-2$$

Rewrite -3x^{2}+10x+8 as \left(-3x^{2}+12x\right)+\left(-2x+8\right).

$$\left(-3x^{2}+12x\right)+\left(-2x+8\right)$$

Factor out 3x in the first and 2 in the second group.

$$3x\left(-x+4\right)+2\left(-x+4\right)$$

Factor out common term -x+4 by using distributive property.

$$\left(-x+4\right)\left(3x+2\right)$$

To find equation solutions, solve -x+4=0 and 3x+2=0.

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

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$x^{2}+6x+9=\left(1-2x\right)^{2}$$

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

$$x^{2}+6x+9=1-4x+4x^{2}$$

Subtract 1 from both sides.

$$x^{2}+6x+9-1=-4x+4x^{2}$$

Subtract 1 from 9 to get 8.

$$x^{2}+6x+8=-4x+4x^{2}$$

Add 4x to both sides.

$$x^{2}+6x+8+4x=4x^{2}$$

Combine 6x and 4x to get 10x.

$$x^{2}+10x+8=4x^{2}$$

Subtract 4x^{2} from both sides.

$$x^{2}+10x+8-4x^{2}=0$$

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

$$-3x^{2}+10x+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 10 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\left(-3\right)\times 8}}{2\left(-3\right)}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\left(-3\right)\times 8}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-10±\sqrt{100+12\times 8}}{2\left(-3\right)}$$

Multiply 12 times 8.

$$x=\frac{-10±\sqrt{100+96}}{2\left(-3\right)}$$

Add 100 to 96.

$$x=\frac{-10±\sqrt{196}}{2\left(-3\right)}$$

Take the square root of 196.

$$x=\frac{-10±14}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-10±14}{-6}$$

Now solve the equation x=\frac{-10±14}{-6} when ± is plus. Add -10 to 14.

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

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

$$x=-\frac{2}{3}$$

Now solve the equation x=\frac{-10±14}{-6} when ± is minus. Subtract 14 from -10.

$$x=-\frac{24}{-6}$$

Divide -24 by -6.

$$x=4$$

The equation is now solved.

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

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$x^{2}+6x+9=\left(1-2x\right)^{2}$$

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

$$x^{2}+6x+9=1-4x+4x^{2}$$

Add 4x to both sides.

$$x^{2}+6x+9+4x=1+4x^{2}$$

Combine 6x and 4x to get 10x.

$$x^{2}+10x+9=1+4x^{2}$$

Subtract 4x^{2} from both sides.

$$x^{2}+10x+9-4x^{2}=1$$

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

$$-3x^{2}+10x+9=1$$

Subtract 9 from both sides.

$$-3x^{2}+10x=1-9$$

Subtract 9 from 1 to get -8.

$$-3x^{2}+10x=-8$$

Divide both sides by -3.

$$\frac{-3x^{2}+10x}{-3}=-\frac{8}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{10}{-3}x=-\frac{8}{-3}$$

Divide 10 by -3.

$$x^{2}-\frac{10}{3}x=-\frac{8}{-3}$$

Divide -8 by -3.

$$x^{2}-\frac{10}{3}x=\frac{8}{3}$$

Divide -\frac{10}{3}, the coefficient of the x term, by 2 to get -\frac{5}{3}. Then add the square of -\frac{5}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{10}{3}x+\left(-\frac{5}{3}\right)^{2}=\frac{8}{3}+\left(-\frac{5}{3}\right)^{2}$$

Square -\frac{5}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{10}{3}x+\frac{25}{9}=\frac{8}{3}+\frac{25}{9}$$

Add \frac{8}{3} to \frac{25}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{10}{3}x+\frac{25}{9}=\frac{49}{9}$$

Factor x^{2}-\frac{10}{3}x+\frac{25}{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}.

$$\left(x-\frac{5}{3}\right)^{2}=\frac{49}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{3}\right)^{2}}=\sqrt{\frac{49}{9}}$$

Simplify.

$$x-\frac{5}{3}=\frac{7}{3} x-\frac{5}{3}=-\frac{7}{3}$$

Add \frac{5}{3} to both sides of the equation.

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