Solve 11*(x-1)^2=(2x-3)^2+4x^2+1

$$11 \cdot ( x - 1 ) ^ { 2 } = ( 2 x - 3 ) ^ { 2 } + 4 x ^ { 2 } + 1$$

Solve for x

$$x = \frac{\sqrt{22} + 5}{3} \approx 3.230138587$$

$$x=\frac{5-\sqrt{22}}{3}\approx 0.103194747$$

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}.

$$11\left(x^{2}-2x+1\right)=\left(2x-3\right)^{2}+4x^{2}+1$$

Use the distributive property to multiply 11 by x^{2}-2x+1.

$$11x^{2}-22x+11=\left(2x-3\right)^{2}+4x^{2}+1$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-3\right)^{2}.

$$11x^{2}-22x+11=4x^{2}-12x+9+4x^{2}+1$$

Combine 4x^{2} and 4x^{2} to get 8x^{2}.

$$11x^{2}-22x+11=8x^{2}-12x+9+1$$

Add 9 and 1 to get 10.

$$11x^{2}-22x+11=8x^{2}-12x+10$$

Subtract 8x^{2} from both sides.

$$11x^{2}-22x+11-8x^{2}=-12x+10$$

Combine 11x^{2} and -8x^{2} to get 3x^{2}.

$$3x^{2}-22x+11=-12x+10$$

Add 12x to both sides.

$$3x^{2}-22x+11+12x=10$$

Combine -22x and 12x to get -10x.

$$3x^{2}-10x+11=10$$

Subtract 10 from both sides.

$$3x^{2}-10x+11-10=0$$

Subtract 10 from 11 to get 1.

$$3x^{2}-10x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -10 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 3}}{2\times 3}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 3}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-10\right)±\sqrt{100-12}}{2\times 3}$$

Add 100 to -12.

$$x=\frac{-\left(-10\right)±\sqrt{88}}{2\times 3}$$

Take the square root of 88.

$$x=\frac{-\left(-10\right)±2\sqrt{22}}{2\times 3}$$

The opposite of -10 is 10.

$$x=\frac{10±2\sqrt{22}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{10±2\sqrt{22}}{6}$$

Now solve the equation x=\frac{10±2\sqrt{22}}{6} when ± is plus. Add 10 to 2\sqrt{22}.

$$x=\frac{2\sqrt{22}+10}{6}$$

Divide 10+2\sqrt{22} by 6.

$$x=\frac{\sqrt{22}+5}{3}$$

Now solve the equation x=\frac{10±2\sqrt{22}}{6} when ± is minus. Subtract 2\sqrt{22} from 10.

$$x=\frac{10-2\sqrt{22}}{6}$$

Divide 10-2\sqrt{22} by 6.

$$x=\frac{5-\sqrt{22}}{3}$$

The equation is now solved.

$$x=\frac{\sqrt{22}+5}{3} x=\frac{5-\sqrt{22}}{3}$$

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}.

$$11\left(x^{2}-2x+1\right)=\left(2x-3\right)^{2}+4x^{2}+1$$

Use the distributive property to multiply 11 by x^{2}-2x+1.

$$11x^{2}-22x+11=\left(2x-3\right)^{2}+4x^{2}+1$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-3\right)^{2}.

$$11x^{2}-22x+11=4x^{2}-12x+9+4x^{2}+1$$

Combine 4x^{2} and 4x^{2} to get 8x^{2}.

$$11x^{2}-22x+11=8x^{2}-12x+9+1$$

Add 9 and 1 to get 10.

$$11x^{2}-22x+11=8x^{2}-12x+10$$

Subtract 8x^{2} from both sides.

$$11x^{2}-22x+11-8x^{2}=-12x+10$$

Combine 11x^{2} and -8x^{2} to get 3x^{2}.

$$3x^{2}-22x+11=-12x+10$$

Add 12x to both sides.

$$3x^{2}-22x+11+12x=10$$

Combine -22x and 12x to get -10x.

$$3x^{2}-10x+11=10$$

Subtract 11 from both sides.

$$3x^{2}-10x=10-11$$

Subtract 11 from 10 to get -1.

$$3x^{2}-10x=-1$$

Divide both sides by 3.

$$\frac{3x^{2}-10x}{3}=-\frac{1}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{10}{3}x=-\frac{1}{3}$$

Divide -\frac{10}{3}, the coefficient of the x term, by 2 to get -\frac{5}{3}. Then add the square of -\frac{5}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{10}{3}x+\left(-\frac{5}{3}\right)^{2}=-\frac{1}{3}+\left(-\frac{5}{3}\right)^{2}$$

Square -\frac{5}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{10}{3}x+\frac{25}{9}=-\frac{1}{3}+\frac{25}{9}$$

Add -\frac{1}{3} to \frac{25}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{10}{3}x+\frac{25}{9}=\frac{22}{9}$$

Factor x^{2}-\frac{10}{3}x+\frac{25}{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}.

$$\left(x-\frac{5}{3}\right)^{2}=\frac{22}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{3}\right)^{2}}=\sqrt{\frac{22}{9}}$$

Simplify.

$$x-\frac{5}{3}=\frac{\sqrt{22}}{3} x-\frac{5}{3}=-\frac{\sqrt{22}}{3}$$

Add \frac{5}{3} to both sides of the equation.

$$x=\frac{\sqrt{22}+5}{3} x=\frac{5-\sqrt{22}}{3}$$