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

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

Solve for x

$$x = -\frac{5}{2} = -2\frac{1}{2} = -2.5$$

$$x = \frac{7}{6} = 1\frac{1}{6} \approx 1.166666667$$

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}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4x-1\right)^{2}.

$$x^{2}-6x+9=\frac{1}{4}\left(16x^{2}-8x+1\right)$$

Use the distributive property to multiply \frac{1}{4} by 16x^{2}-8x+1.

$$x^{2}-6x+9=4x^{2}-2x+\frac{1}{4}$$

Subtract 4x^{2} from both sides.

$$x^{2}-6x+9-4x^{2}=-2x+\frac{1}{4}$$

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

$$-3x^{2}-6x+9=-2x+\frac{1}{4}$$

Add 2x to both sides.

$$-3x^{2}-6x+9+2x=\frac{1}{4}$$

Combine -6x and 2x to get -4x.

$$-3x^{2}-4x+9=\frac{1}{4}$$

Subtract \frac{1}{4} from both sides.

$$-3x^{2}-4x+9-\frac{1}{4}=0$$

Subtract \frac{1}{4} from 9 to get \frac{35}{4}.

$$-3x^{2}-4x+\frac{35}{4}=0$$

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

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-3\right)\times \frac{35}{4}}}{2\left(-3\right)}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-3\right)\times \frac{35}{4}}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-\left(-4\right)±\sqrt{16+12\times \frac{35}{4}}}{2\left(-3\right)}$$

Multiply 12 times \frac{35}{4}.

$$x=\frac{-\left(-4\right)±\sqrt{16+105}}{2\left(-3\right)}$$

Add 16 to 105.

$$x=\frac{-\left(-4\right)±\sqrt{121}}{2\left(-3\right)}$$

Take the square root of 121.

$$x=\frac{-\left(-4\right)±11}{2\left(-3\right)}$$

The opposite of -4 is 4.

$$x=\frac{4±11}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{4±11}{-6}$$

Now solve the equation x=\frac{4±11}{-6} when ± is plus. Add 4 to 11.

$$x=\frac{15}{-6}$$

Reduce the fraction \frac{15}{-6} to lowest terms by extracting and canceling out 3.

$$x=-\frac{5}{2}$$

Now solve the equation x=\frac{4±11}{-6} when ± is minus. Subtract 11 from 4.

$$x=-\frac{7}{-6}$$

Divide -7 by -6.

$$x=\frac{7}{6}$$

The equation is now solved.

$$x=-\frac{5}{2} x=\frac{7}{6}$$

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}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4x-1\right)^{2}.

$$x^{2}-6x+9=\frac{1}{4}\left(16x^{2}-8x+1\right)$$

Use the distributive property to multiply \frac{1}{4} by 16x^{2}-8x+1.

$$x^{2}-6x+9=4x^{2}-2x+\frac{1}{4}$$

Subtract 4x^{2} from both sides.

$$x^{2}-6x+9-4x^{2}=-2x+\frac{1}{4}$$

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

$$-3x^{2}-6x+9=-2x+\frac{1}{4}$$

Add 2x to both sides.

$$-3x^{2}-6x+9+2x=\frac{1}{4}$$

Combine -6x and 2x to get -4x.

$$-3x^{2}-4x+9=\frac{1}{4}$$

Subtract 9 from both sides.

$$-3x^{2}-4x=\frac{1}{4}-9$$

Subtract 9 from \frac{1}{4} to get -\frac{35}{4}.

$$-3x^{2}-4x=-\frac{35}{4}$$

Divide both sides by -3.

$$\frac{-3x^{2}-4x}{-3}=-\frac{\frac{35}{4}}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\left(-\frac{4}{-3}\right)x=-\frac{\frac{35}{4}}{-3}$$

Divide -4 by -3.

$$x^{2}+\frac{4}{3}x=-\frac{\frac{35}{4}}{-3}$$

Divide -\frac{35}{4} by -3.

$$x^{2}+\frac{4}{3}x=\frac{35}{12}$$

Divide \frac{4}{3}, the coefficient of the x term, by 2 to get \frac{2}{3}. Then add the square of \frac{2}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{4}{3}x+\left(\frac{2}{3}\right)^{2}=\frac{35}{12}+\left(\frac{2}{3}\right)^{2}$$

Square \frac{2}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=\frac{35}{12}+\frac{4}{9}$$

Add \frac{35}{12} to \frac{4}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=\frac{121}{36}$$

Factor x^{2}+\frac{4}{3}x+\frac{4}{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}.

$$\left(x+\frac{2}{3}\right)^{2}=\frac{121}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{2}{3}\right)^{2}}=\sqrt{\frac{121}{36}}$$

Simplify.

$$x+\frac{2}{3}=\frac{11}{6} x+\frac{2}{3}=-\frac{11}{6}$$

Subtract \frac{2}{3} from both sides of the equation.

$$x=\frac{7}{6} x=-\frac{5}{2}$$