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

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

Solve for x

$$x = \frac{\sqrt{13} + 5}{3} \approx 2.868517092$$

$$x=\frac{5-\sqrt{13}}{3}\approx 0.464816242$$

Steps Using the Quadratic Formula

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

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

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

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

Add 3 and 1 to get 4.

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

Subtract 4x from both sides.

$$4-6x+3x^{2}-4x=0$$

Combine -6x and -4x to get -10x.

$$4-10x+3x^{2}=0$$

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.

$$3x^{2}-10x+4=0$$

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

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

Square -10.

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

Multiply -4 times 3.

$$x=\frac{-\left(-10\right)±\sqrt{100-12\times 4}}{2\times 3}$$

Multiply -12 times 4.

$$x=\frac{-\left(-10\right)±\sqrt{100-48}}{2\times 3}$$

Add 100 to -48.

$$x=\frac{-\left(-10\right)±\sqrt{52}}{2\times 3}$$

Take the square root of 52.

$$x=\frac{-\left(-10\right)±2\sqrt{13}}{2\times 3}$$

The opposite of -10 is 10.

$$x=\frac{10±2\sqrt{13}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{10±2\sqrt{13}}{6}$$

Now solve the equation x=\frac{10±2\sqrt{13}}{6} when ± is plus. Add 10 to 2\sqrt{13}.

$$x=\frac{2\sqrt{13}+10}{6}$$

Divide 10+2\sqrt{13} by 6.

$$x=\frac{\sqrt{13}+5}{3}$$

Now solve the equation x=\frac{10±2\sqrt{13}}{6} when ± is minus. Subtract 2\sqrt{13} from 10.

$$x=\frac{10-2\sqrt{13}}{6}$$

Divide 10-2\sqrt{13} by 6.

$$x=\frac{5-\sqrt{13}}{3}$$

The equation is now solved.

$$x=\frac{\sqrt{13}+5}{3} x=\frac{5-\sqrt{13}}{3}$$

Steps for Completing the Square

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

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

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

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

Add 3 and 1 to get 4.

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

Subtract 4x from both sides.

$$4-6x+3x^{2}-4x=0$$

Combine -6x and -4x to get -10x.

$$4-10x+3x^{2}=0$$

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

$$-10x+3x^{2}=-4$$

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.

$$3x^{2}-10x=-4$$

Divide both sides by 3.

$$\frac{3x^{2}-10x}{3}=-\frac{4}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{10}{3}x=-\frac{4}{3}$$

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.

$$x^{2}-\frac{10}{3}x+\left(-\frac{5}{3}\right)^{2}=-\frac{4}{3}+\left(-\frac{5}{3}\right)^{2}$$

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

$$x^{2}-\frac{10}{3}x+\frac{25}{9}=-\frac{4}{3}+\frac{25}{9}$$

Add -\frac{4}{3} to \frac{25}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{10}{3}x+\frac{25}{9}=\frac{13}{9}$$

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}.

$$\left(x-\frac{5}{3}\right)^{2}=\frac{13}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{3}\right)^{2}}=\sqrt{\frac{13}{9}}$$

Simplify.

$$x-\frac{5}{3}=\frac{\sqrt{13}}{3} x-\frac{5}{3}=-\frac{\sqrt{13}}{3}$$

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

$$x=\frac{\sqrt{13}+5}{3} x=\frac{5-\sqrt{13}}{3}$$