Solve (x-1)^2/4-3x+1/8=2

$$\frac { ( x - 1 ) ^ { 2 } } { 4 } - \frac { 3 x + 1 } { 8 } = 2$$

Solve for x

$$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

$$x=5$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 8, the least common multiple of 4,8.

$$2\left(x-1\right)^{2}-\left(3x+1\right)=16$$

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

$$2\left(x^{2}-2x+1\right)-\left(3x+1\right)=16$$

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

$$2x^{2}-4x+2-\left(3x+1\right)=16$$

To find the opposite of 3x+1, find the opposite of each term.

$$2x^{2}-4x+2-3x-1=16$$

Combine -4x and -3x to get -7x.

$$2x^{2}-7x+2-1=16$$

Subtract 1 from 2 to get 1.

$$2x^{2}-7x+1=16$$

Subtract 16 from both sides.

$$2x^{2}-7x+1-16=0$$

Subtract 16 from 1 to get -15.

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

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

$$a+b=-7 ab=2\left(-15\right)=-30$$

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 -30.

$$1,-30 2,-15 3,-10 5,-6$$

Calculate the sum for each pair.

$$1-30=-29 2-15=-13 3-10=-7 5-6=-1$$

The solution is the pair that gives sum -7.

$$a=-10 b=3$$

Rewrite 2x^{2}-7x-15 as \left(2x^{2}-10x\right)+\left(3x-15\right).

$$\left(2x^{2}-10x\right)+\left(3x-15\right)$$

Factor out 2x in the first and 3 in the second group.

$$2x\left(x-5\right)+3\left(x-5\right)$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(2x+3\right)$$

To find equation solutions, solve x-5=0 and 2x+3=0.

$$x=5 x=-\frac{3}{2}$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 8, the least common multiple of 4,8.

$$2\left(x-1\right)^{2}-\left(3x+1\right)=16$$

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

$$2\left(x^{2}-2x+1\right)-\left(3x+1\right)=16$$

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

$$2x^{2}-4x+2-\left(3x+1\right)=16$$

To find the opposite of 3x+1, find the opposite of each term.

$$2x^{2}-4x+2-3x-1=16$$

Combine -4x and -3x to get -7x.

$$2x^{2}-7x+2-1=16$$

Subtract 1 from 2 to get 1.

$$2x^{2}-7x+1=16$$

Subtract 16 from both sides.

$$2x^{2}-7x+1-16=0$$

Subtract 16 from 1 to get -15.

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

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

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

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 2\left(-15\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-7\right)±\sqrt{49-8\left(-15\right)}}{2\times 2}$$

Multiply -8 times -15.

$$x=\frac{-\left(-7\right)±\sqrt{49+120}}{2\times 2}$$

Add 49 to 120.

$$x=\frac{-\left(-7\right)±\sqrt{169}}{2\times 2}$$

Take the square root of 169.

$$x=\frac{-\left(-7\right)±13}{2\times 2}$$

The opposite of -7 is 7.

$$x=\frac{7±13}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{7±13}{4}$$

Now solve the equation x=\frac{7±13}{4} when ± is plus. Add 7 to 13.

$$x=\frac{20}{4}$$

Divide 20 by 4.

$$x=5$$

Now solve the equation x=\frac{7±13}{4} when ± is minus. Subtract 13 from 7.

$$x=-\frac{6}{4}$$

Reduce the fraction \frac{-6}{4} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

$$x=5 x=-\frac{3}{2}$$

Steps for Completing the Square

Multiply both sides of the equation by 8, the least common multiple of 4,8.

$$2\left(x-1\right)^{2}-\left(3x+1\right)=16$$

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

$$2\left(x^{2}-2x+1\right)-\left(3x+1\right)=16$$

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

$$2x^{2}-4x+2-\left(3x+1\right)=16$$

To find the opposite of 3x+1, find the opposite of each term.

$$2x^{2}-4x+2-3x-1=16$$

Combine -4x and -3x to get -7x.

$$2x^{2}-7x+2-1=16$$

Subtract 1 from 2 to get 1.

$$2x^{2}-7x+1=16$$

Subtract 1 from both sides.

$$2x^{2}-7x=16-1$$

Subtract 1 from 16 to get 15.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide -\frac{7}{2}, the coefficient of the x term, by 2 to get -\frac{7}{4}. Then add the square of -\frac{7}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{2}x+\left(-\frac{7}{4}\right)^{2}=\frac{15}{2}+\left(-\frac{7}{4}\right)^{2}$$

Square -\frac{7}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=\frac{15}{2}+\frac{49}{16}$$

Add \frac{15}{2} to \frac{49}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=\frac{169}{16}$$

Factor x^{2}-\frac{7}{2}x+\frac{49}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{7}{4}\right)^{2}=\frac{169}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{4}\right)^{2}}=\sqrt{\frac{169}{16}}$$

Simplify.

$$x-\frac{7}{4}=\frac{13}{4} x-\frac{7}{4}=-\frac{13}{4}$$

Add \frac{7}{4} to both sides of the equation.

$$x=5 x=-\frac{3}{2}$$