Solve 4(x+3)^2=(x-1)^2

$$4 ( x + 3 ) ^ { 2 } = ( x - 1 ) ^ { 2 }$$

Solve for x

$$x=-7$$

$$x = -\frac{5}{3} = -1\frac{2}{3} \approx -1.666666667$$

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}.

$$4\left(x^{2}+6x+9\right)=\left(x-1\right)^{2}$$

Use the distributive property to multiply 4 by x^{2}+6x+9.

$$4x^{2}+24x+36=\left(x-1\right)^{2}$$

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

$$4x^{2}+24x+36=x^{2}-2x+1$$

Subtract x^{2} from both sides.

$$4x^{2}+24x+36-x^{2}=-2x+1$$

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

$$3x^{2}+24x+36=-2x+1$$

Add 2x to both sides.

$$3x^{2}+24x+36+2x=1$$

Combine 24x and 2x to get 26x.

$$3x^{2}+26x+36=1$$

Subtract 1 from both sides.

$$3x^{2}+26x+36-1=0$$

Subtract 1 from 36 to get 35.

$$3x^{2}+26x+35=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx+35. To find a and b, set up a system to be solved.

$$a+b=26 ab=3\times 35=105$$

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 105.

$$1,105 3,35 5,21 7,15$$

Calculate the sum for each pair.

$$1+105=106 3+35=38 5+21=26 7+15=22$$

The solution is the pair that gives sum 26.

$$a=5 b=21$$

Rewrite 3x^{2}+26x+35 as \left(3x^{2}+5x\right)+\left(21x+35\right).

$$\left(3x^{2}+5x\right)+\left(21x+35\right)$$

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

$$x\left(3x+5\right)+7\left(3x+5\right)$$

Factor out common term 3x+5 by using distributive property.

$$\left(3x+5\right)\left(x+7\right)$$

To find equation solutions, solve 3x+5=0 and x+7=0.

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

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}.

$$4\left(x^{2}+6x+9\right)=\left(x-1\right)^{2}$$

Use the distributive property to multiply 4 by x^{2}+6x+9.

$$4x^{2}+24x+36=\left(x-1\right)^{2}$$

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

$$4x^{2}+24x+36=x^{2}-2x+1$$

Subtract x^{2} from both sides.

$$4x^{2}+24x+36-x^{2}=-2x+1$$

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

$$3x^{2}+24x+36=-2x+1$$

Add 2x to both sides.

$$3x^{2}+24x+36+2x=1$$

Combine 24x and 2x to get 26x.

$$3x^{2}+26x+36=1$$

Subtract 1 from both sides.

$$3x^{2}+26x+36-1=0$$

Subtract 1 from 36 to get 35.

$$3x^{2}+26x+35=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 26 for b, and 35 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-26±\sqrt{26^{2}-4\times 3\times 35}}{2\times 3}$$

Square 26.

$$x=\frac{-26±\sqrt{676-4\times 3\times 35}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-26±\sqrt{676-12\times 35}}{2\times 3}$$

Multiply -12 times 35.

$$x=\frac{-26±\sqrt{676-420}}{2\times 3}$$

Add 676 to -420.

$$x=\frac{-26±\sqrt{256}}{2\times 3}$$

Take the square root of 256.

$$x=\frac{-26±16}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-26±16}{6}$$

Now solve the equation x=\frac{-26±16}{6} when ± is plus. Add -26 to 16.

$$x=-\frac{10}{6}$$

Reduce the fraction \frac{-10}{6} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{-26±16}{6} when ± is minus. Subtract 16 from -26.

$$x=-\frac{42}{6}$$

Divide -42 by 6.

$$x=-7$$

The equation is now solved.

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

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}.

$$4\left(x^{2}+6x+9\right)=\left(x-1\right)^{2}$$

Use the distributive property to multiply 4 by x^{2}+6x+9.

$$4x^{2}+24x+36=\left(x-1\right)^{2}$$

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

$$4x^{2}+24x+36=x^{2}-2x+1$$

Subtract x^{2} from both sides.

$$4x^{2}+24x+36-x^{2}=-2x+1$$

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

$$3x^{2}+24x+36=-2x+1$$

Add 2x to both sides.

$$3x^{2}+24x+36+2x=1$$

Combine 24x and 2x to get 26x.

$$3x^{2}+26x+36=1$$

Subtract 36 from both sides.

$$3x^{2}+26x=1-36$$

Subtract 36 from 1 to get -35.

$$3x^{2}+26x=-35$$

Divide both sides by 3.

$$\frac{3x^{2}+26x}{3}=-\frac{35}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{26}{3}x=-\frac{35}{3}$$

Divide \frac{26}{3}, the coefficient of the x term, by 2 to get \frac{13}{3}. Then add the square of \frac{13}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{26}{3}x+\left(\frac{13}{3}\right)^{2}=-\frac{35}{3}+\left(\frac{13}{3}\right)^{2}$$

Square \frac{13}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{26}{3}x+\frac{169}{9}=-\frac{35}{3}+\frac{169}{9}$$

Add -\frac{35}{3} to \frac{169}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{26}{3}x+\frac{169}{9}=\frac{64}{9}$$

Factor x^{2}+\frac{26}{3}x+\frac{169}{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}.

$$\left(x+\frac{13}{3}\right)^{2}=\frac{64}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{13}{3}\right)^{2}}=\sqrt{\frac{64}{9}}$$

Simplify.

$$x+\frac{13}{3}=\frac{8}{3} x+\frac{13}{3}=-\frac{8}{3}$$

Subtract \frac{13}{3} from both sides of the equation.

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