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

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

Solve for x

$$x=\sqrt{10}+5\approx 8.16227766$$

$$x=5-\sqrt{10}\approx 1.83772234$$

Steps Using the Quadratic Formula

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

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

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

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

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

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

Use the distributive property to multiply \frac{3}{2} by x^{2}+2x+1.

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

Combine \frac{1}{2}x^{2} and \frac{3}{2}x^{2} to get 2x^{2}.

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

Combine -x and 3x to get 2x.

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

Add \frac{1}{2} and \frac{3}{2} to get 2.

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

Consider \left(2x-1\right)\left(2x+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.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

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

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

Use the distributive property to multiply -3 by x^{2}-4x+4.

$$2x^{2}+2x+2=4x^{2}-1-3x^{2}+12x-12$$

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

$$2x^{2}+2x+2=x^{2}-1+12x-12$$

Subtract 12 from -1 to get -13.

$$2x^{2}+2x+2=x^{2}-13+12x$$

Subtract x^{2} from both sides.

$$2x^{2}+2x+2-x^{2}=-13+12x$$

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

$$x^{2}+2x+2=-13+12x$$

Subtract -13 from both sides.

$$x^{2}+2x+2-\left(-13\right)=12x$$

The opposite of -13 is 13.

$$x^{2}+2x+2+13=12x$$

Subtract 12x from both sides.

$$x^{2}+2x+2+13-12x=0$$

Add 2 and 13 to get 15.

$$x^{2}+2x+15-12x=0$$

Combine 2x and -12x to get -10x.

$$x^{2}-10x+15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and 15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 15}}{2}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 15}}{2}$$

Multiply -4 times 15.

$$x=\frac{-\left(-10\right)±\sqrt{100-60}}{2}$$

Add 100 to -60.

$$x=\frac{-\left(-10\right)±\sqrt{40}}{2}$$

Take the square root of 40.

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

The opposite of -10 is 10.

$$x=\frac{10±2\sqrt{10}}{2}$$

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

$$x=\frac{2\sqrt{10}+10}{2}$$

Divide 10+2\sqrt{10} by 2.

$$x=\sqrt{10}+5$$

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

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

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

$$x=5-\sqrt{10}$$

The equation is now solved.

$$x=\sqrt{10}+5 x=5-\sqrt{10}$$

Steps for Completing the Square

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

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

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

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

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

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

Use the distributive property to multiply \frac{3}{2} by x^{2}+2x+1.

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

Combine \frac{1}{2}x^{2} and \frac{3}{2}x^{2} to get 2x^{2}.

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

Combine -x and 3x to get 2x.

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

Add \frac{1}{2} and \frac{3}{2} to get 2.

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

Consider \left(2x-1\right)\left(2x+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.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

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

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

Use the distributive property to multiply -3 by x^{2}-4x+4.

$$2x^{2}+2x+2=4x^{2}-1-3x^{2}+12x-12$$

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

$$2x^{2}+2x+2=x^{2}-1+12x-12$$

Subtract 12 from -1 to get -13.

$$2x^{2}+2x+2=x^{2}-13+12x$$

Subtract x^{2} from both sides.

$$2x^{2}+2x+2-x^{2}=-13+12x$$

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

$$x^{2}+2x+2=-13+12x$$

Subtract 12x from both sides.

$$x^{2}+2x+2-12x=-13$$

Combine 2x and -12x to get -10x.

$$x^{2}-10x+2=-13$$

Subtract 2 from both sides.

$$x^{2}-10x=-13-2$$

Subtract 2 from -13 to get -15.

$$x^{2}-10x=-15$$

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

$$x^{2}-10x+\left(-5\right)^{2}=-15+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=-15+25$$

Add -15 to 25.

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

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

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{10}$$

Simplify.

$$x-5=\sqrt{10} x-5=-\sqrt{10}$$

Add 5 to both sides of the equation.

$$x=\sqrt{10}+5 x=5-\sqrt{10}$$