Solve x=(2-1.1x)^2

$$x = ( 2 - 1.1 x ) ^ { 2 }$$

Solve for x

$$x = \frac{70 \sqrt{5} + 270}{121} \approx 3.524998004$$

$$x=\frac{270-70\sqrt{5}}{121}\approx 0.937811914$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2-1.1x\right)^{2}.

$$x=4-4.4x+1.21x^{2}$$

Subtract 4 from both sides.

$$x-4=-4.4x+1.21x^{2}$$

Add 4.4x to both sides.

$$x-4+4.4x=1.21x^{2}$$

Combine x and 4.4x to get 5.4x.

$$5.4x-4=1.21x^{2}$$

Subtract 1.21x^{2} from both sides.

$$5.4x-4-1.21x^{2}=0$$

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.

$$-1.21x^{2}+5.4x-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1.21 for a, 5.4 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5.4±\sqrt{5.4^{2}-4\left(-1.21\right)\left(-4\right)}}{2\left(-1.21\right)}$$

Square 5.4 by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-5.4±\sqrt{29.16-4\left(-1.21\right)\left(-4\right)}}{2\left(-1.21\right)}$$

Multiply -4 times -1.21.

$$x=\frac{-5.4±\sqrt{29.16+4.84\left(-4\right)}}{2\left(-1.21\right)}$$

Multiply 4.84 times -4.

$$x=\frac{-5.4±\sqrt{\frac{729-484}{25}}}{2\left(-1.21\right)}$$

Add 29.16 to -19.36 by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-5.4±\sqrt{9.8}}{2\left(-1.21\right)}$$

Take the square root of 9.8.

$$x=\frac{-5.4±\frac{7\sqrt{5}}{5}}{2\left(-1.21\right)}$$

Multiply 2 times -1.21.

$$x=\frac{-5.4±\frac{7\sqrt{5}}{5}}{-2.42}$$

Now solve the equation x=\frac{-5.4±\frac{7\sqrt{5}}{5}}{-2.42} when ± is plus. Add -5.4 to \frac{7\sqrt{5}}{5}.

$$x=\frac{7\sqrt{5}-27}{-2.42\times 5}$$

Divide \frac{-27+7\sqrt{5}}{5} by -2.42 by multiplying \frac{-27+7\sqrt{5}}{5} by the reciprocal of -2.42.

$$x=\frac{270-70\sqrt{5}}{121}$$

Now solve the equation x=\frac{-5.4±\frac{7\sqrt{5}}{5}}{-2.42} when ± is minus. Subtract \frac{7\sqrt{5}}{5} from -5.4.

$$x=\frac{-7\sqrt{5}-27}{-2.42\times 5}$$

Divide \frac{-27-7\sqrt{5}}{5} by -2.42 by multiplying \frac{-27-7\sqrt{5}}{5} by the reciprocal of -2.42.

$$x=\frac{70\sqrt{5}+270}{121}$$

The equation is now solved.

$$x=\frac{270-70\sqrt{5}}{121} x=\frac{70\sqrt{5}+270}{121}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2-1.1x\right)^{2}.

$$x=4-4.4x+1.21x^{2}$$

Add 4.4x to both sides.

$$x+4.4x=4+1.21x^{2}$$

Combine x and 4.4x to get 5.4x.

$$5.4x=4+1.21x^{2}$$

Subtract 1.21x^{2} from both sides.

$$5.4x-1.21x^{2}=4$$

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.

$$-1.21x^{2}+5.4x=4$$

Divide both sides of the equation by -1.21, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-1.21x^{2}+5.4x}{-1.21}=\frac{4}{-1.21}$$

Dividing by -1.21 undoes the multiplication by -1.21.

$$x^{2}+\frac{5.4}{-1.21}x=\frac{4}{-1.21}$$

Divide 5.4 by -1.21 by multiplying 5.4 by the reciprocal of -1.21.

$$x^{2}-\frac{540}{121}x=\frac{4}{-1.21}$$

Divide 4 by -1.21 by multiplying 4 by the reciprocal of -1.21.

$$x^{2}-\frac{540}{121}x=-\frac{400}{121}$$

Divide -\frac{540}{121}, the coefficient of the x term, by 2 to get -\frac{270}{121}. Then add the square of -\frac{270}{121} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{540}{121}x+\left(-\frac{270}{121}\right)^{2}=-\frac{400}{121}+\left(-\frac{270}{121}\right)^{2}$$

Square -\frac{270}{121} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{540}{121}x+\frac{72900}{14641}=-\frac{400}{121}+\frac{72900}{14641}$$

Add -\frac{400}{121} to \frac{72900}{14641} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{540}{121}x+\frac{72900}{14641}=\frac{24500}{14641}$$

Factor x^{2}-\frac{540}{121}x+\frac{72900}{14641}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{270}{121}\right)^{2}=\frac{24500}{14641}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{270}{121}\right)^{2}}=\sqrt{\frac{24500}{14641}}$$

Simplify.

$$x-\frac{270}{121}=\frac{70\sqrt{5}}{121} x-\frac{270}{121}=-\frac{70\sqrt{5}}{121}$$

Add \frac{270}{121} to both sides of the equation.

$$x=\frac{70\sqrt{5}+270}{121} x=\frac{270-70\sqrt{5}}{121}$$