Solve x^2-4left(2x-1right)^2+x+3left(2x-1right)=1

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

Solve for x

$$x=1$$

$$x=\frac{8}{15}\approx 0.533333333$$

Steps Using Factoring By Grouping

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

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

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

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

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

$$-15x^{2}+16x-4+x+3\left(2x-1\right)=1$$

Combine 16x and x to get 17x.

$$-15x^{2}+17x-4+3\left(2x-1\right)=1$$

Use the distributive property to multiply 3 by 2x-1.

$$-15x^{2}+17x-4+6x-3=1$$

Combine 17x and 6x to get 23x.

$$-15x^{2}+23x-4-3=1$$

Subtract 3 from -4 to get -7.

$$-15x^{2}+23x-7=1$$

Subtract 1 from both sides.

$$-15x^{2}+23x-7-1=0$$

Subtract 1 from -7 to get -8.

$$-15x^{2}+23x-8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -15x^{2}+ax+bx-8. To find a and b, set up a system to be solved.

$$a+b=23 ab=-15\left(-8\right)=120$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 120.

$$1,120 2,60 3,40 4,30 5,24 6,20 8,15 10,12$$

Calculate the sum for each pair.

$$1+120=121 2+60=62 3+40=43 4+30=34 5+24=29 6+20=26 8+15=23 10+12=22$$

The solution is the pair that gives sum 23.

$$a=15 b=8$$

Rewrite -15x^{2}+23x-8 as \left(-15x^{2}+15x\right)+\left(8x-8\right).

$$\left(-15x^{2}+15x\right)+\left(8x-8\right)$$

Factor out 15x in the first and -8 in the second group.

$$15x\left(-x+1\right)-8\left(-x+1\right)$$

Factor out common term -x+1 by using distributive property.

$$\left(-x+1\right)\left(15x-8\right)$$

To find equation solutions, solve -x+1=0 and 15x-8=0.

$$x=1 x=\frac{8}{15}$$

Steps Using the Quadratic Formula

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

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

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

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

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

$$-15x^{2}+16x-4+x+3\left(2x-1\right)=1$$

Combine 16x and x to get 17x.

$$-15x^{2}+17x-4+3\left(2x-1\right)=1$$

Use the distributive property to multiply 3 by 2x-1.

$$-15x^{2}+17x-4+6x-3=1$$

Combine 17x and 6x to get 23x.

$$-15x^{2}+23x-4-3=1$$

Subtract 3 from -4 to get -7.

$$-15x^{2}+23x-7=1$$

Subtract 1 from both sides.

$$-15x^{2}+23x-7-1=0$$

Subtract 1 from -7 to get -8.

$$-15x^{2}+23x-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -15 for a, 23 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-23±\sqrt{23^{2}-4\left(-15\right)\left(-8\right)}}{2\left(-15\right)}$$

Square 23.

$$x=\frac{-23±\sqrt{529-4\left(-15\right)\left(-8\right)}}{2\left(-15\right)}$$

Multiply -4 times -15.

$$x=\frac{-23±\sqrt{529+60\left(-8\right)}}{2\left(-15\right)}$$

Multiply 60 times -8.

$$x=\frac{-23±\sqrt{529-480}}{2\left(-15\right)}$$

Add 529 to -480.

$$x=\frac{-23±\sqrt{49}}{2\left(-15\right)}$$

Take the square root of 49.

$$x=\frac{-23±7}{2\left(-15\right)}$$

Multiply 2 times -15.

$$x=\frac{-23±7}{-30}$$

Now solve the equation x=\frac{-23±7}{-30} when ± is plus. Add -23 to 7.

$$x=-\frac{16}{-30}$$

Reduce the fraction \frac{-16}{-30} to lowest terms by extracting and canceling out 2.

$$x=\frac{8}{15}$$

Now solve the equation x=\frac{-23±7}{-30} when ± is minus. Subtract 7 from -23.

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

Divide -30 by -30.

$$x=1$$

The equation is now solved.

$$x=\frac{8}{15} x=1$$

Steps for Completing the Square

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

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

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

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

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

$$-15x^{2}+16x-4+x+3\left(2x-1\right)=1$$

Combine 16x and x to get 17x.

$$-15x^{2}+17x-4+3\left(2x-1\right)=1$$

Use the distributive property to multiply 3 by 2x-1.

$$-15x^{2}+17x-4+6x-3=1$$

Combine 17x and 6x to get 23x.

$$-15x^{2}+23x-4-3=1$$

Subtract 3 from -4 to get -7.

$$-15x^{2}+23x-7=1$$

Add 7 to both sides.

$$-15x^{2}+23x=1+7$$

Add 1 and 7 to get 8.

$$-15x^{2}+23x=8$$

Divide both sides by -15.

$$\frac{-15x^{2}+23x}{-15}=\frac{8}{-15}$$

Dividing by -15 undoes the multiplication by -15.

$$x^{2}+\frac{23}{-15}x=\frac{8}{-15}$$

Divide 23 by -15.

$$x^{2}-\frac{23}{15}x=\frac{8}{-15}$$

Divide 8 by -15.

$$x^{2}-\frac{23}{15}x=-\frac{8}{15}$$

Divide -\frac{23}{15}, the coefficient of the x term, by 2 to get -\frac{23}{30}. Then add the square of -\frac{23}{30} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{23}{15}x+\left(-\frac{23}{30}\right)^{2}=-\frac{8}{15}+\left(-\frac{23}{30}\right)^{2}$$

Square -\frac{23}{30} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{23}{15}x+\frac{529}{900}=-\frac{8}{15}+\frac{529}{900}$$

Add -\frac{8}{15} to \frac{529}{900} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{23}{15}x+\frac{529}{900}=\frac{49}{900}$$

Factor x^{2}-\frac{23}{15}x+\frac{529}{900}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{23}{30}\right)^{2}=\frac{49}{900}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{23}{30}\right)^{2}}=\sqrt{\frac{49}{900}}$$

Simplify.

$$x-\frac{23}{30}=\frac{7}{30} x-\frac{23}{30}=-\frac{7}{30}$$

Add \frac{23}{30} to both sides of the equation.

$$x=1 x=\frac{8}{15}$$