Solve left(3x-7right)^2-4left(x+1right)^2=0

$${ \left(3x-7 \right) }^{ 2 } -4 { \left(x+1 \right) }^{ 2 } =0$$

Solve for x

$$x=1$$

$$x=9$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-7\right)^{2}.

$$9x^{2}-42x+49-4\left(x+1\right)^{2}=0$$

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

$$9x^{2}-42x+49-4\left(x^{2}+2x+1\right)=0$$

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

$$9x^{2}-42x+49-4x^{2}-8x-4=0$$

Combine 9x^{2} and -4x^{2} to get 5x^{2}.

$$5x^{2}-42x+49-8x-4=0$$

Combine -42x and -8x to get -50x.

$$5x^{2}-50x+49-4=0$$

Subtract 4 from 49 to get 45.

$$5x^{2}-50x+45=0$$

Divide both sides by 5.

$$x^{2}-10x+9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+9. To find a and b, set up a system to be solved.

$$a+b=-10 ab=1\times 9=9$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 9.

$$-1,-9 -3,-3$$

Calculate the sum for each pair.

$$-1-9=-10 -3-3=-6$$

The solution is the pair that gives sum -10.

$$a=-9 b=-1$$

Rewrite x^{2}-10x+9 as \left(x^{2}-9x\right)+\left(-x+9\right).

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

Factor out x in the first and -1 in the second group.

$$x\left(x-9\right)-\left(x-9\right)$$

Factor out common term x-9 by using distributive property.

$$\left(x-9\right)\left(x-1\right)$$

To find equation solutions, solve x-9=0 and x-1=0.

$$x=9 x=1$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-7\right)^{2}.

$$9x^{2}-42x+49-4\left(x+1\right)^{2}=0$$

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

$$9x^{2}-42x+49-4\left(x^{2}+2x+1\right)=0$$

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

$$9x^{2}-42x+49-4x^{2}-8x-4=0$$

Combine 9x^{2} and -4x^{2} to get 5x^{2}.

$$5x^{2}-42x+49-8x-4=0$$

Combine -42x and -8x to get -50x.

$$5x^{2}-50x+49-4=0$$

Subtract 4 from 49 to get 45.

$$5x^{2}-50x+45=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -50 for b, and 45 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-50\right)±\sqrt{\left(-50\right)^{2}-4\times 5\times 45}}{2\times 5}$$

Square -50.

$$x=\frac{-\left(-50\right)±\sqrt{2500-4\times 5\times 45}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-50\right)±\sqrt{2500-20\times 45}}{2\times 5}$$

Multiply -20 times 45.

$$x=\frac{-\left(-50\right)±\sqrt{2500-900}}{2\times 5}$$

Add 2500 to -900.

$$x=\frac{-\left(-50\right)±\sqrt{1600}}{2\times 5}$$

Take the square root of 1600.

$$x=\frac{-\left(-50\right)±40}{2\times 5}$$

The opposite of -50 is 50.

$$x=\frac{50±40}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{50±40}{10}$$

Now solve the equation x=\frac{50±40}{10} when ± is plus. Add 50 to 40.

$$x=\frac{90}{10}$$

Divide 90 by 10.

$$x=9$$

Now solve the equation x=\frac{50±40}{10} when ± is minus. Subtract 40 from 50.

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

Divide 10 by 10.

$$x=1$$

The equation is now solved.

$$x=9 x=1$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-7\right)^{2}.

$$9x^{2}-42x+49-4\left(x+1\right)^{2}=0$$

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

$$9x^{2}-42x+49-4\left(x^{2}+2x+1\right)=0$$

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

$$9x^{2}-42x+49-4x^{2}-8x-4=0$$

Combine 9x^{2} and -4x^{2} to get 5x^{2}.

$$5x^{2}-42x+49-8x-4=0$$

Combine -42x and -8x to get -50x.

$$5x^{2}-50x+49-4=0$$

Subtract 4 from 49 to get 45.

$$5x^{2}-50x+45=0$$

Subtract 45 from both sides. Anything subtracted from zero gives its negation.

$$5x^{2}-50x=-45$$

Divide both sides by 5.

$$\frac{5x^{2}-50x}{5}=-\frac{45}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}+\left(-\frac{50}{5}\right)x=-\frac{45}{5}$$

Divide -50 by 5.

$$x^{2}-10x=-\frac{45}{5}$$

Divide -45 by 5.

$$x^{2}-10x=-9$$

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

$$x^{2}-10x+\left(-5\right)^{2}=-9+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=-9+25$$

Add -9 to 25.

$$x^{2}-10x+25=16$$

Factor x^{2}-10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-5\right)^{2}=16$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{16}$$

Simplify.

$$x-5=4 x-5=-4$$

Add 5 to both sides of the equation.

$$x=9 x=1$$