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

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

Solve for x

$$x=-6$$

$$x=\frac{2}{3}\approx 0.666666667$$

Steps Using Factoring By Grouping

Variable x cannot be equal to 4 since division by zero is not defined. Multiply both sides of the equation by 4\left(x-4\right)^{2}, the least common multiple of \left(4-x\right)^{2}+\left(-1+x-3\right)^{2},4.

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

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

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

Subtract 3 from 4 to get 1.

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

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

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

Add 1 and 1 to get 2.

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

Combine 2x and 2x to get 4x.

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

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

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

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

$$4+8x+4x^{2}=\left(x-4\right)^{2}$$

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

$$4+8x+4x^{2}=x^{2}-8x+16$$

Subtract x^{2} from both sides.

$$4+8x+4x^{2}-x^{2}=-8x+16$$

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

$$4+8x+3x^{2}=-8x+16$$

Add 8x to both sides.

$$4+8x+3x^{2}+8x=16$$

Combine 8x and 8x to get 16x.

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

Subtract 16 from both sides.

$$4+16x+3x^{2}-16=0$$

Subtract 16 from 4 to get -12.

$$-12+16x+3x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$3x^{2}+16x-12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-12. To find a and b, set up a system to be solved.

$$a+b=16 ab=3\left(-12\right)=-36$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -36.

$$-1,36 -2,18 -3,12 -4,9 -6,6$$

Calculate the sum for each pair.

$$-1+36=35 -2+18=16 -3+12=9 -4+9=5 -6+6=0$$

The solution is the pair that gives sum 16.

$$a=-2 b=18$$

Rewrite 3x^{2}+16x-12 as \left(3x^{2}-2x\right)+\left(18x-12\right).

$$\left(3x^{2}-2x\right)+\left(18x-12\right)$$

Factor out x in the first and 6 in the second group.

$$x\left(3x-2\right)+6\left(3x-2\right)$$

Factor out common term 3x-2 by using distributive property.

$$\left(3x-2\right)\left(x+6\right)$$

To find equation solutions, solve 3x-2=0 and x+6=0.

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

Steps Using the Quadratic Formula

Variable x cannot be equal to 4 since division by zero is not defined. Multiply both sides of the equation by 4\left(x-4\right)^{2}, the least common multiple of \left(4-x\right)^{2}+\left(-1+x-3\right)^{2},4.

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

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

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

Subtract 3 from 4 to get 1.

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

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

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

Add 1 and 1 to get 2.

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

Combine 2x and 2x to get 4x.

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

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

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

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

$$4+8x+4x^{2}=\left(x-4\right)^{2}$$

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

$$4+8x+4x^{2}=x^{2}-8x+16$$

Subtract x^{2} from both sides.

$$4+8x+4x^{2}-x^{2}=-8x+16$$

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

$$4+8x+3x^{2}=-8x+16$$

Add 8x to both sides.

$$4+8x+3x^{2}+8x=16$$

Combine 8x and 8x to get 16x.

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

Subtract 16 from both sides.

$$4+16x+3x^{2}-16=0$$

Subtract 16 from 4 to get -12.

$$-12+16x+3x^{2}=0$$

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.

$$3x^{2}+16x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 16 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 16.

$$x=\frac{-16±\sqrt{256-4\times 3\left(-12\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-16±\sqrt{256-12\left(-12\right)}}{2\times 3}$$

Multiply -12 times -12.

$$x=\frac{-16±\sqrt{256+144}}{2\times 3}$$

Add 256 to 144.

$$x=\frac{-16±\sqrt{400}}{2\times 3}$$

Take the square root of 400.

$$x=\frac{-16±20}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-16±20}{6}$$

Now solve the equation x=\frac{-16±20}{6} when ± is plus. Add -16 to 20.

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

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

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

Now solve the equation x=\frac{-16±20}{6} when ± is minus. Subtract 20 from -16.

$$x=-\frac{36}{6}$$

Divide -36 by 6.

$$x=-6$$

The equation is now solved.

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

Steps for Completing the Square

Variable x cannot be equal to 4 since division by zero is not defined. Multiply both sides of the equation by 4\left(x-4\right)^{2}, the least common multiple of \left(4-x\right)^{2}+\left(-1+x-3\right)^{2},4.

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

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

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

Subtract 3 from 4 to get 1.

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

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

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

Add 1 and 1 to get 2.

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

Combine 2x and 2x to get 4x.

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

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

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

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

$$4+8x+4x^{2}=\left(x-4\right)^{2}$$

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

$$4+8x+4x^{2}=x^{2}-8x+16$$

Subtract x^{2} from both sides.

$$4+8x+4x^{2}-x^{2}=-8x+16$$

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

$$4+8x+3x^{2}=-8x+16$$

Add 8x to both sides.

$$4+8x+3x^{2}+8x=16$$

Combine 8x and 8x to get 16x.

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

Subtract 4 from both sides.

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

Subtract 4 from 16 to get 12.

$$16x+3x^{2}=12$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$3x^{2}+16x=12$$

Divide both sides by 3.

$$\frac{3x^{2}+16x}{3}=\frac{12}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{16}{3}x=\frac{12}{3}$$

Divide 12 by 3.

$$x^{2}+\frac{16}{3}x=4$$

Divide \frac{16}{3}, the coefficient of the x term, by 2 to get \frac{8}{3}. Then add the square of \frac{8}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{16}{3}x+\left(\frac{8}{3}\right)^{2}=4+\left(\frac{8}{3}\right)^{2}$$

Square \frac{8}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{16}{3}x+\frac{64}{9}=4+\frac{64}{9}$$

Add 4 to \frac{64}{9}.

$$x^{2}+\frac{16}{3}x+\frac{64}{9}=\frac{100}{9}$$

Factor x^{2}+\frac{16}{3}x+\frac{64}{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}.

$$\left(x+\frac{8}{3}\right)^{2}=\frac{100}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{8}{3}\right)^{2}}=\sqrt{\frac{100}{9}}$$

Simplify.

$$x+\frac{8}{3}=\frac{10}{3} x+\frac{8}{3}=-\frac{10}{3}$$

Subtract \frac{8}{3} from both sides of the equation.

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