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

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

Solve for x

$$x=-\frac{1}{5}=-0.2$$

$$x = \frac{7}{3} = 2\frac{1}{3} \approx 2.333333333$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4x-3\right)^{2}.

$$16x^{2}-24x+9=\left(x+4\right)^{2}$$

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

$$16x^{2}-24x+9=x^{2}+8x+16$$

Subtract x^{2} from both sides.

$$16x^{2}-24x+9-x^{2}=8x+16$$

Combine 16x^{2} and -x^{2} to get 15x^{2}.

$$15x^{2}-24x+9=8x+16$$

Subtract 8x from both sides.

$$15x^{2}-24x+9-8x=16$$

Combine -24x and -8x to get -32x.

$$15x^{2}-32x+9=16$$

Subtract 16 from both sides.

$$15x^{2}-32x+9-16=0$$

Subtract 16 from 9 to get -7.

$$15x^{2}-32x-7=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 15x^{2}+ax+bx-7. To find a and b, set up a system to be solved.

$$a+b=-32 ab=15\left(-7\right)=-105$$

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 -105.

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

Calculate the sum for each pair.

$$1-105=-104 3-35=-32 5-21=-16 7-15=-8$$

The solution is the pair that gives sum -32.

$$a=-35 b=3$$

Rewrite 15x^{2}-32x-7 as \left(15x^{2}-35x\right)+\left(3x-7\right).

$$\left(15x^{2}-35x\right)+\left(3x-7\right)$$

Factor out 5x in 15x^{2}-35x.

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

Factor out common term 3x-7 by using distributive property.

$$\left(3x-7\right)\left(5x+1\right)$$

To find equation solutions, solve 3x-7=0 and 5x+1=0.

$$x=\frac{7}{3} x=-\frac{1}{5}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4x-3\right)^{2}.

$$16x^{2}-24x+9=\left(x+4\right)^{2}$$

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

$$16x^{2}-24x+9=x^{2}+8x+16$$

Subtract x^{2} from both sides.

$$16x^{2}-24x+9-x^{2}=8x+16$$

Combine 16x^{2} and -x^{2} to get 15x^{2}.

$$15x^{2}-24x+9=8x+16$$

Subtract 8x from both sides.

$$15x^{2}-24x+9-8x=16$$

Combine -24x and -8x to get -32x.

$$15x^{2}-32x+9=16$$

Subtract 16 from both sides.

$$15x^{2}-32x+9-16=0$$

Subtract 16 from 9 to get -7.

$$15x^{2}-32x-7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 15 for a, -32 for b, and -7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-32\right)±\sqrt{\left(-32\right)^{2}-4\times 15\left(-7\right)}}{2\times 15}$$

Square -32.

$$x=\frac{-\left(-32\right)±\sqrt{1024-4\times 15\left(-7\right)}}{2\times 15}$$

Multiply -4 times 15.

$$x=\frac{-\left(-32\right)±\sqrt{1024-60\left(-7\right)}}{2\times 15}$$

Multiply -60 times -7.

$$x=\frac{-\left(-32\right)±\sqrt{1024+420}}{2\times 15}$$

Add 1024 to 420.

$$x=\frac{-\left(-32\right)±\sqrt{1444}}{2\times 15}$$

Take the square root of 1444.

$$x=\frac{-\left(-32\right)±38}{2\times 15}$$

The opposite of -32 is 32.

$$x=\frac{32±38}{2\times 15}$$

Multiply 2 times 15.

$$x=\frac{32±38}{30}$$

Now solve the equation x=\frac{32±38}{30} when ± is plus. Add 32 to 38.

$$x=\frac{70}{30}$$

Reduce the fraction \frac{70}{30} to lowest terms by extracting and canceling out 10.

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

Now solve the equation x=\frac{32±38}{30} when ± is minus. Subtract 38 from 32.

$$x=-\frac{6}{30}$$

Reduce the fraction \frac{-6}{30} to lowest terms by extracting and canceling out 6.

$$x=-\frac{1}{5}$$

The equation is now solved.

$$x=\frac{7}{3} x=-\frac{1}{5}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4x-3\right)^{2}.

$$16x^{2}-24x+9=\left(x+4\right)^{2}$$

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

$$16x^{2}-24x+9=x^{2}+8x+16$$

Subtract x^{2} from both sides.

$$16x^{2}-24x+9-x^{2}=8x+16$$

Combine 16x^{2} and -x^{2} to get 15x^{2}.

$$15x^{2}-24x+9=8x+16$$

Subtract 8x from both sides.

$$15x^{2}-24x+9-8x=16$$

Combine -24x and -8x to get -32x.

$$15x^{2}-32x+9=16$$

Subtract 9 from both sides.

$$15x^{2}-32x=16-9$$

Subtract 9 from 16 to get 7.

$$15x^{2}-32x=7$$

Divide both sides by 15.

$$\frac{15x^{2}-32x}{15}=\frac{7}{15}$$

Dividing by 15 undoes the multiplication by 15.

$$x^{2}-\frac{32}{15}x=\frac{7}{15}$$

Divide -\frac{32}{15}, the coefficient of the x term, by 2 to get -\frac{16}{15}. Then add the square of -\frac{16}{15} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{32}{15}x+\left(-\frac{16}{15}\right)^{2}=\frac{7}{15}+\left(-\frac{16}{15}\right)^{2}$$

Square -\frac{16}{15} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{32}{15}x+\frac{256}{225}=\frac{7}{15}+\frac{256}{225}$$

Add \frac{7}{15} to \frac{256}{225} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{32}{15}x+\frac{256}{225}=\frac{361}{225}$$

Factor x^{2}-\frac{32}{15}x+\frac{256}{225}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{16}{15}\right)^{2}=\frac{361}{225}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{16}{15}\right)^{2}}=\sqrt{\frac{361}{225}}$$

Simplify.

$$x-\frac{16}{15}=\frac{19}{15} x-\frac{16}{15}=-\frac{19}{15}$$

Add \frac{16}{15} to both sides of the equation.

$$x=\frac{7}{3} x=-\frac{1}{5}$$