Solve (4x-1)(4x+1)=(5x-1)^2

$$( 4 x - 1 ) ( 4 x + 1 ) = ( 5 x - 1 ) ^ { 2 }$$

Solve for x

$$x=\frac{\sqrt{7}+5}{9}\approx 0.849527923$$

$$x=\frac{5-\sqrt{7}}{9}\approx 0.261583188$$

Steps Using the Quadratic Formula

Consider \left(4x-1\right)\left(4x+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$\left(4x\right)^{2}-1=\left(5x-1\right)^{2}$$

Expand \left(4x\right)^{2}.

$$4^{2}x^{2}-1=\left(5x-1\right)^{2}$$

Calculate 4 to the power of 2 and get 16.

$$16x^{2}-1=\left(5x-1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5x-1\right)^{2}.

$$16x^{2}-1=25x^{2}-10x+1$$

Subtract 25x^{2} from both sides.

$$16x^{2}-1-25x^{2}=-10x+1$$

Combine 16x^{2} and -25x^{2} to get -9x^{2}.

$$-9x^{2}-1=-10x+1$$

Add 10x to both sides.

$$-9x^{2}-1+10x=1$$

Subtract 1 from both sides.

$$-9x^{2}-1+10x-1=0$$

Subtract 1 from -1 to get -2.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -9 for a, 10 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\left(-9\right)\left(-2\right)}}{2\left(-9\right)}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\left(-9\right)\left(-2\right)}}{2\left(-9\right)}$$

Multiply -4 times -9.

$$x=\frac{-10±\sqrt{100+36\left(-2\right)}}{2\left(-9\right)}$$

Multiply 36 times -2.

$$x=\frac{-10±\sqrt{100-72}}{2\left(-9\right)}$$

Add 100 to -72.

$$x=\frac{-10±\sqrt{28}}{2\left(-9\right)}$$

Take the square root of 28.

$$x=\frac{-10±2\sqrt{7}}{2\left(-9\right)}$$

Multiply 2 times -9.

$$x=\frac{-10±2\sqrt{7}}{-18}$$

Now solve the equation x=\frac{-10±2\sqrt{7}}{-18} when ± is plus. Add -10 to 2\sqrt{7}.

$$x=\frac{2\sqrt{7}-10}{-18}$$

Divide -10+2\sqrt{7} by -18.

$$x=\frac{5-\sqrt{7}}{9}$$

Now solve the equation x=\frac{-10±2\sqrt{7}}{-18} when ± is minus. Subtract 2\sqrt{7} from -10.

$$x=\frac{-2\sqrt{7}-10}{-18}$$

Divide -10-2\sqrt{7} by -18.

$$x=\frac{\sqrt{7}+5}{9}$$

The equation is now solved.

$$x=\frac{5-\sqrt{7}}{9} x=\frac{\sqrt{7}+5}{9}$$

Steps for Completing the Square

Consider \left(4x-1\right)\left(4x+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$\left(4x\right)^{2}-1=\left(5x-1\right)^{2}$$

Expand \left(4x\right)^{2}.

$$4^{2}x^{2}-1=\left(5x-1\right)^{2}$$

Calculate 4 to the power of 2 and get 16.

$$16x^{2}-1=\left(5x-1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5x-1\right)^{2}.

$$16x^{2}-1=25x^{2}-10x+1$$

Subtract 25x^{2} from both sides.

$$16x^{2}-1-25x^{2}=-10x+1$$

Combine 16x^{2} and -25x^{2} to get -9x^{2}.

$$-9x^{2}-1=-10x+1$$

Add 10x to both sides.

$$-9x^{2}-1+10x=1$$

Add 1 to both sides.

$$-9x^{2}+10x=1+1$$

Add 1 and 1 to get 2.

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

Divide both sides by -9.

$$\frac{-9x^{2}+10x}{-9}=\frac{2}{-9}$$

Dividing by -9 undoes the multiplication by -9.

$$x^{2}+\frac{10}{-9}x=\frac{2}{-9}$$

Divide 10 by -9.

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

Divide 2 by -9.

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

Divide -\frac{10}{9}, the coefficient of the x term, by 2 to get -\frac{5}{9}. Then add the square of -\frac{5}{9} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{5}{9} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{10}{9}x+\frac{25}{81}=-\frac{2}{9}+\frac{25}{81}$$

Add -\frac{2}{9} to \frac{25}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{10}{9}x+\frac{25}{81}=\frac{7}{81}$$

Factor x^{2}-\frac{10}{9}x+\frac{25}{81}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{9}\right)^{2}=\frac{7}{81}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{9}\right)^{2}}=\sqrt{\frac{7}{81}}$$

Simplify.

$$x-\frac{5}{9}=\frac{\sqrt{7}}{9} x-\frac{5}{9}=-\frac{\sqrt{7}}{9}$$

Add \frac{5}{9} to both sides of the equation.

$$x=\frac{\sqrt{7}+5}{9} x=\frac{5-\sqrt{7}}{9}$$