Solve left(8x-5right)^2=left(4x-3right)^2

$${ \left(8x-5 \right) }^{ 2 } = { \left(4x-3 \right) }^{ 2 }$$

Solve for x

$$x=\frac{1}{2}=0.5$$

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

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(8x-5\right)^{2}.

$$64x^{2}-80x+25=\left(4x-3\right)^{2}$$

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

$$64x^{2}-80x+25=16x^{2}-24x+9$$

Subtract 16x^{2} from both sides.

$$64x^{2}-80x+25-16x^{2}=-24x+9$$

Combine 64x^{2} and -16x^{2} to get 48x^{2}.

$$48x^{2}-80x+25=-24x+9$$

Add 24x to both sides.

$$48x^{2}-80x+25+24x=9$$

Combine -80x and 24x to get -56x.

$$48x^{2}-56x+25=9$$

Subtract 9 from both sides.

$$48x^{2}-56x+25-9=0$$

Subtract 9 from 25 to get 16.

$$48x^{2}-56x+16=0$$

Divide both sides by 8.

$$6x^{2}-7x+2=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 6x^{2}+ax+bx+2. To find a and b, set up a system to be solved.

$$a+b=-7 ab=6\times 2=12$$

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 12.

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

Calculate the sum for each pair.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

The solution is the pair that gives sum -7.

$$a=-4 b=-3$$

Rewrite 6x^{2}-7x+2 as \left(6x^{2}-4x\right)+\left(-3x+2\right).

$$\left(6x^{2}-4x\right)+\left(-3x+2\right)$$

Factor out 2x in the first and -1 in the second group.

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

Factor out common term 3x-2 by using distributive property.

$$\left(3x-2\right)\left(2x-1\right)$$

To find equation solutions, solve 3x-2=0 and 2x-1=0.

$$x=\frac{2}{3} x=\frac{1}{2}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(8x-5\right)^{2}.

$$64x^{2}-80x+25=\left(4x-3\right)^{2}$$

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

$$64x^{2}-80x+25=16x^{2}-24x+9$$

Subtract 16x^{2} from both sides.

$$64x^{2}-80x+25-16x^{2}=-24x+9$$

Combine 64x^{2} and -16x^{2} to get 48x^{2}.

$$48x^{2}-80x+25=-24x+9$$

Add 24x to both sides.

$$48x^{2}-80x+25+24x=9$$

Combine -80x and 24x to get -56x.

$$48x^{2}-56x+25=9$$

Subtract 9 from both sides.

$$48x^{2}-56x+25-9=0$$

Subtract 9 from 25 to get 16.

$$48x^{2}-56x+16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 48 for a, -56 for b, and 16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-56\right)±\sqrt{\left(-56\right)^{2}-4\times 48\times 16}}{2\times 48}$$

Square -56.

$$x=\frac{-\left(-56\right)±\sqrt{3136-4\times 48\times 16}}{2\times 48}$$

Multiply -4 times 48.

$$x=\frac{-\left(-56\right)±\sqrt{3136-192\times 16}}{2\times 48}$$

Multiply -192 times 16.

$$x=\frac{-\left(-56\right)±\sqrt{3136-3072}}{2\times 48}$$

Add 3136 to -3072.

$$x=\frac{-\left(-56\right)±\sqrt{64}}{2\times 48}$$

Take the square root of 64.

$$x=\frac{-\left(-56\right)±8}{2\times 48}$$

The opposite of -56 is 56.

$$x=\frac{56±8}{2\times 48}$$

Multiply 2 times 48.

$$x=\frac{56±8}{96}$$

Now solve the equation x=\frac{56±8}{96} when ± is plus. Add 56 to 8.

$$x=\frac{64}{96}$$

Reduce the fraction \frac{64}{96} to lowest terms by extracting and canceling out 32.

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

Now solve the equation x=\frac{56±8}{96} when ± is minus. Subtract 8 from 56.

$$x=\frac{48}{96}$$

Reduce the fraction \frac{48}{96} to lowest terms by extracting and canceling out 48.

$$x=\frac{1}{2}$$

The equation is now solved.

$$x=\frac{2}{3} x=\frac{1}{2}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(8x-5\right)^{2}.

$$64x^{2}-80x+25=\left(4x-3\right)^{2}$$

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

$$64x^{2}-80x+25=16x^{2}-24x+9$$

Subtract 16x^{2} from both sides.

$$64x^{2}-80x+25-16x^{2}=-24x+9$$

Combine 64x^{2} and -16x^{2} to get 48x^{2}.

$$48x^{2}-80x+25=-24x+9$$

Add 24x to both sides.

$$48x^{2}-80x+25+24x=9$$

Combine -80x and 24x to get -56x.

$$48x^{2}-56x+25=9$$

Subtract 25 from both sides.

$$48x^{2}-56x=9-25$$

Subtract 25 from 9 to get -16.

$$48x^{2}-56x=-16$$

Divide both sides by 48.

$$\frac{48x^{2}-56x}{48}=-\frac{16}{48}$$

Dividing by 48 undoes the multiplication by 48.

$$x^{2}+\left(-\frac{56}{48}\right)x=-\frac{16}{48}$$

Reduce the fraction \frac{-56}{48} to lowest terms by extracting and canceling out 8.

$$x^{2}-\frac{7}{6}x=-\frac{16}{48}$$

Reduce the fraction \frac{-16}{48} to lowest terms by extracting and canceling out 16.

$$x^{2}-\frac{7}{6}x=-\frac{1}{3}$$

Divide -\frac{7}{6}, the coefficient of the x term, by 2 to get -\frac{7}{12}. Then add the square of -\frac{7}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{6}x+\left(-\frac{7}{12}\right)^{2}=-\frac{1}{3}+\left(-\frac{7}{12}\right)^{2}$$

Square -\frac{7}{12} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{6}x+\frac{49}{144}=-\frac{1}{3}+\frac{49}{144}$$

Add -\frac{1}{3} to \frac{49}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{7}{6}x+\frac{49}{144}=\frac{1}{144}$$

Factor x^{2}-\frac{7}{6}x+\frac{49}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{7}{12}\right)^{2}=\frac{1}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{12}\right)^{2}}=\sqrt{\frac{1}{144}}$$

Simplify.

$$x-\frac{7}{12}=\frac{1}{12} x-\frac{7}{12}=-\frac{1}{12}$$

Add \frac{7}{12} to both sides of the equation.

$$x=\frac{2}{3} x=\frac{1}{2}$$