Solve (-4x+1)^2=9(x+2)^2

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

Solve for x

$$x=7$$

$$x=-\frac{5}{7}\approx -0.714285714$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-4x+1\right)^{2}.

$$16x^{2}-8x+1=9\left(x+2\right)^{2}$$

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

$$16x^{2}-8x+1=9\left(x^{2}+4x+4\right)$$

Use the distributive property to multiply 9 by x^{2}+4x+4.

$$16x^{2}-8x+1=9x^{2}+36x+36$$

Subtract 9x^{2} from both sides.

$$16x^{2}-8x+1-9x^{2}=36x+36$$

Combine 16x^{2} and -9x^{2} to get 7x^{2}.

$$7x^{2}-8x+1=36x+36$$

Subtract 36x from both sides.

$$7x^{2}-8x+1-36x=36$$

Combine -8x and -36x to get -44x.

$$7x^{2}-44x+1=36$$

Subtract 36 from both sides.

$$7x^{2}-44x+1-36=0$$

Subtract 36 from 1 to get -35.

$$7x^{2}-44x-35=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 7x^{2}+ax+bx-35. To find a and b, set up a system to be solved.

$$a+b=-44 ab=7\left(-35\right)=-245$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -245.

$$1,-245 5,-49 7,-35$$

Calculate the sum for each pair.

$$1-245=-244 5-49=-44 7-35=-28$$

The solution is the pair that gives sum -44.

$$a=-49 b=5$$

Rewrite 7x^{2}-44x-35 as \left(7x^{2}-49x\right)+\left(5x-35\right).

$$\left(7x^{2}-49x\right)+\left(5x-35\right)$$

Factor out 7x in the first and 5 in the second group.

$$7x\left(x-7\right)+5\left(x-7\right)$$

Factor out common term x-7 by using distributive property.

$$\left(x-7\right)\left(7x+5\right)$$

To find equation solutions, solve x-7=0 and 7x+5=0.

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

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-4x+1\right)^{2}.

$$16x^{2}-8x+1=9\left(x+2\right)^{2}$$

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

$$16x^{2}-8x+1=9\left(x^{2}+4x+4\right)$$

Use the distributive property to multiply 9 by x^{2}+4x+4.

$$16x^{2}-8x+1=9x^{2}+36x+36$$

Subtract 9x^{2} from both sides.

$$16x^{2}-8x+1-9x^{2}=36x+36$$

Combine 16x^{2} and -9x^{2} to get 7x^{2}.

$$7x^{2}-8x+1=36x+36$$

Subtract 36x from both sides.

$$7x^{2}-8x+1-36x=36$$

Combine -8x and -36x to get -44x.

$$7x^{2}-44x+1=36$$

Subtract 36 from both sides.

$$7x^{2}-44x+1-36=0$$

Subtract 36 from 1 to get -35.

$$7x^{2}-44x-35=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 7 for a, -44 for b, and -35 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-44\right)±\sqrt{\left(-44\right)^{2}-4\times 7\left(-35\right)}}{2\times 7}$$

Square -44.

$$x=\frac{-\left(-44\right)±\sqrt{1936-4\times 7\left(-35\right)}}{2\times 7}$$

Multiply -4 times 7.

$$x=\frac{-\left(-44\right)±\sqrt{1936-28\left(-35\right)}}{2\times 7}$$

Multiply -28 times -35.

$$x=\frac{-\left(-44\right)±\sqrt{1936+980}}{2\times 7}$$

Add 1936 to 980.

$$x=\frac{-\left(-44\right)±\sqrt{2916}}{2\times 7}$$

Take the square root of 2916.

$$x=\frac{-\left(-44\right)±54}{2\times 7}$$

The opposite of -44 is 44.

$$x=\frac{44±54}{2\times 7}$$

Multiply 2 times 7.

$$x=\frac{44±54}{14}$$

Now solve the equation x=\frac{44±54}{14} when ± is plus. Add 44 to 54.

$$x=\frac{98}{14}$$

Divide 98 by 14.

$$x=7$$

Now solve the equation x=\frac{44±54}{14} when ± is minus. Subtract 54 from 44.

$$x=-\frac{10}{14}$$

Reduce the fraction \frac{-10}{14} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

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

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-4x+1\right)^{2}.

$$16x^{2}-8x+1=9\left(x+2\right)^{2}$$

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

$$16x^{2}-8x+1=9\left(x^{2}+4x+4\right)$$

Use the distributive property to multiply 9 by x^{2}+4x+4.

$$16x^{2}-8x+1=9x^{2}+36x+36$$

Subtract 9x^{2} from both sides.

$$16x^{2}-8x+1-9x^{2}=36x+36$$

Combine 16x^{2} and -9x^{2} to get 7x^{2}.

$$7x^{2}-8x+1=36x+36$$

Subtract 36x from both sides.

$$7x^{2}-8x+1-36x=36$$

Combine -8x and -36x to get -44x.

$$7x^{2}-44x+1=36$$

Subtract 1 from both sides.

$$7x^{2}-44x=36-1$$

Subtract 1 from 36 to get 35.

$$7x^{2}-44x=35$$

Divide both sides by 7.

$$\frac{7x^{2}-44x}{7}=\frac{35}{7}$$

Dividing by 7 undoes the multiplication by 7.

$$x^{2}-\frac{44}{7}x=\frac{35}{7}$$

Divide 35 by 7.

$$x^{2}-\frac{44}{7}x=5$$

Divide -\frac{44}{7}, the coefficient of the x term, by 2 to get -\frac{22}{7}. Then add the square of -\frac{22}{7} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{44}{7}x+\left(-\frac{22}{7}\right)^{2}=5+\left(-\frac{22}{7}\right)^{2}$$

Square -\frac{22}{7} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{44}{7}x+\frac{484}{49}=5+\frac{484}{49}$$

Add 5 to \frac{484}{49}.

$$x^{2}-\frac{44}{7}x+\frac{484}{49}=\frac{729}{49}$$

Factor x^{2}-\frac{44}{7}x+\frac{484}{49}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{22}{7}\right)^{2}=\frac{729}{49}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{22}{7}\right)^{2}}=\sqrt{\frac{729}{49}}$$

Simplify.

$$x-\frac{22}{7}=\frac{27}{7} x-\frac{22}{7}=-\frac{27}{7}$$

Add \frac{22}{7} to both sides of the equation.

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