Solve x^2+5x-1=4x^2+x+1/3

$$x ^ { 2 } + 5 x - 1 = 4 x ^ { 2 } + x + \frac { 1 } { 3 }$$

Solve for x

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

Steps Using the Quadratic Formula

Subtract 4x^{2} from both sides.

$$x^{2}+5x-1-4x^{2}=x+\frac{1}{3}$$

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

$$-3x^{2}+5x-1=x+\frac{1}{3}$$

Subtract x from both sides.

$$-3x^{2}+5x-1-x=\frac{1}{3}$$

Combine 5x and -x to get 4x.

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

Subtract \frac{1}{3} from both sides.

$$-3x^{2}+4x-1-\frac{1}{3}=0$$

Subtract \frac{1}{3} from -1 to get -\frac{4}{3}.

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

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

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

Square 4.

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

Multiply -4 times -3.

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

Multiply 12 times -\frac{4}{3}.

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

Add 16 to -16.

$$x=\frac{-4±\sqrt{0}}{2\left(-3\right)}$$

Take the square root of 0.

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

Multiply 2 times -3.

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

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

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

Steps for Completing the Square

Subtract 4x^{2} from both sides.

$$x^{2}+5x-1-4x^{2}=x+\frac{1}{3}$$

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

$$-3x^{2}+5x-1=x+\frac{1}{3}$$

Subtract x from both sides.

$$-3x^{2}+5x-1-x=\frac{1}{3}$$

Combine 5x and -x to get 4x.

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

Add 1 to both sides.

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

Add \frac{1}{3} and 1 to get \frac{4}{3}.

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

Divide both sides by -3.

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

Dividing by -3 undoes the multiplication by -3.

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

Divide 4 by -3.

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

Divide \frac{4}{3} by -3.

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

Divide -\frac{4}{3}, the coefficient of the x term, by 2 to get -\frac{2}{3}. Then add the square of -\frac{2}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{4}{3}x+\left(-\frac{2}{3}\right)^{2}=-\frac{4}{9}+\left(-\frac{2}{3}\right)^{2}$$

Square -\frac{2}{3} by squaring both the numerator and the denominator of the fraction.

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

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

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=0$$

Factor x^{2}-\frac{4}{3}x+\frac{4}{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}.

$$\left(x-\frac{2}{3}\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{3}\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-\frac{2}{3}=0 x-\frac{2}{3}=0$$

Add \frac{2}{3} to both sides of the equation.

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

The equation is now solved. Solutions are the same.

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