Solve n+4/n^2-3n+1quadn=3

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

Solve for n

$$n = \frac{\sqrt{145} + 13}{4} \approx 6.260398645$$

$$n=\frac{13-\sqrt{145}}{4}\approx 0.239601355$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by \left(n-\left(-\frac{1}{2}\sqrt{5}+\frac{3}{2}\right)\right)\left(n-\left(\frac{1}{2}\sqrt{5}+\frac{3}{2}\right)\right).

$$\left(n+4\right)n=3\left(n-\left(-\frac{1}{2}\sqrt{5}+\frac{3}{2}\right)\right)\left(n-\left(\frac{1}{2}\sqrt{5}+\frac{3}{2}\right)\right)$$

Use the distributive property to multiply n+4 by n.

$$n^{2}+4n=3\left(n-\left(-\frac{1}{2}\sqrt{5}+\frac{3}{2}\right)\right)\left(n-\left(\frac{1}{2}\sqrt{5}+\frac{3}{2}\right)\right)$$

To find the opposite of -\frac{1}{2}\sqrt{5}+\frac{3}{2}, find the opposite of each term.

$$n^{2}+4n=3\left(n+\frac{1}{2}\sqrt{5}-\frac{3}{2}\right)\left(n-\left(\frac{1}{2}\sqrt{5}+\frac{3}{2}\right)\right)$$

To find the opposite of \frac{1}{2}\sqrt{5}+\frac{3}{2}, find the opposite of each term.

$$n^{2}+4n=3\left(n+\frac{1}{2}\sqrt{5}-\frac{3}{2}\right)\left(n-\frac{1}{2}\sqrt{5}-\frac{3}{2}\right)$$

Use the distributive property to multiply 3 by n+\frac{1}{2}\sqrt{5}-\frac{3}{2}.

$$n^{2}+4n=\left(3n+\frac{3}{2}\sqrt{5}-\frac{9}{2}\right)\left(n-\frac{1}{2}\sqrt{5}-\frac{3}{2}\right)$$

Use the distributive property to multiply 3n+\frac{3}{2}\sqrt{5}-\frac{9}{2} by n-\frac{1}{2}\sqrt{5}-\frac{3}{2} and combine like terms.

$$n^{2}+4n=3n^{2}-9n-\frac{3}{4}\left(\sqrt{5}\right)^{2}+\frac{27}{4}$$

The square of \sqrt{5} is 5.

$$n^{2}+4n=3n^{2}-9n-\frac{3}{4}\times 5+\frac{27}{4}$$

Multiply -\frac{3}{4} and 5 to get -\frac{15}{4}.

$$n^{2}+4n=3n^{2}-9n-\frac{15}{4}+\frac{27}{4}$$

Add -\frac{15}{4} and \frac{27}{4} to get 3.

$$n^{2}+4n=3n^{2}-9n+3$$

Subtract 3n^{2} from both sides.

$$n^{2}+4n-3n^{2}=-9n+3$$

Combine n^{2} and -3n^{2} to get -2n^{2}.

$$-2n^{2}+4n=-9n+3$$

Add 9n to both sides.

$$-2n^{2}+4n+9n=3$$

Combine 4n and 9n to get 13n.

$$-2n^{2}+13n=3$$

Subtract 3 from both sides.

$$-2n^{2}+13n-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 13 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-13±\sqrt{13^{2}-4\left(-2\right)\left(-3\right)}}{2\left(-2\right)}$$

Square 13.

$$n=\frac{-13±\sqrt{169-4\left(-2\right)\left(-3\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$n=\frac{-13±\sqrt{169+8\left(-3\right)}}{2\left(-2\right)}$$

Multiply 8 times -3.

$$n=\frac{-13±\sqrt{169-24}}{2\left(-2\right)}$$

Add 169 to -24.

$$n=\frac{-13±\sqrt{145}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$n=\frac{-13±\sqrt{145}}{-4}$$

Now solve the equation n=\frac{-13±\sqrt{145}}{-4} when ± is plus. Add -13 to \sqrt{145}.

$$n=\frac{\sqrt{145}-13}{-4}$$

Divide -13+\sqrt{145} by -4.

$$n=\frac{13-\sqrt{145}}{4}$$

Now solve the equation n=\frac{-13±\sqrt{145}}{-4} when ± is minus. Subtract \sqrt{145} from -13.

$$n=\frac{-\sqrt{145}-13}{-4}$$

Divide -13-\sqrt{145} by -4.

$$n=\frac{\sqrt{145}+13}{4}$$

The equation is now solved.

$$n=\frac{13-\sqrt{145}}{4} n=\frac{\sqrt{145}+13}{4}$$

Steps for Completing the Square

Multiply both sides of the equation by \left(n-\left(-\frac{1}{2}\sqrt{5}+\frac{3}{2}\right)\right)\left(n-\left(\frac{1}{2}\sqrt{5}+\frac{3}{2}\right)\right).

$$\left(n+4\right)n=3\left(n-\left(-\frac{1}{2}\sqrt{5}+\frac{3}{2}\right)\right)\left(n-\left(\frac{1}{2}\sqrt{5}+\frac{3}{2}\right)\right)$$

Use the distributive property to multiply n+4 by n.

$$n^{2}+4n=3\left(n-\left(-\frac{1}{2}\sqrt{5}+\frac{3}{2}\right)\right)\left(n-\left(\frac{1}{2}\sqrt{5}+\frac{3}{2}\right)\right)$$

To find the opposite of -\frac{1}{2}\sqrt{5}+\frac{3}{2}, find the opposite of each term.

$$n^{2}+4n=3\left(n+\frac{1}{2}\sqrt{5}-\frac{3}{2}\right)\left(n-\left(\frac{1}{2}\sqrt{5}+\frac{3}{2}\right)\right)$$

To find the opposite of \frac{1}{2}\sqrt{5}+\frac{3}{2}, find the opposite of each term.

$$n^{2}+4n=3\left(n+\frac{1}{2}\sqrt{5}-\frac{3}{2}\right)\left(n-\frac{1}{2}\sqrt{5}-\frac{3}{2}\right)$$

Use the distributive property to multiply 3 by n+\frac{1}{2}\sqrt{5}-\frac{3}{2}.

$$n^{2}+4n=\left(3n+\frac{3}{2}\sqrt{5}-\frac{9}{2}\right)\left(n-\frac{1}{2}\sqrt{5}-\frac{3}{2}\right)$$

Use the distributive property to multiply 3n+\frac{3}{2}\sqrt{5}-\frac{9}{2} by n-\frac{1}{2}\sqrt{5}-\frac{3}{2} and combine like terms.

$$n^{2}+4n=3n^{2}-9n-\frac{3}{4}\left(\sqrt{5}\right)^{2}+\frac{27}{4}$$

The square of \sqrt{5} is 5.

$$n^{2}+4n=3n^{2}-9n-\frac{3}{4}\times 5+\frac{27}{4}$$

Multiply -\frac{3}{4} and 5 to get -\frac{15}{4}.

$$n^{2}+4n=3n^{2}-9n-\frac{15}{4}+\frac{27}{4}$$

Add -\frac{15}{4} and \frac{27}{4} to get 3.

$$n^{2}+4n=3n^{2}-9n+3$$

Subtract 3n^{2} from both sides.

$$n^{2}+4n-3n^{2}=-9n+3$$

Combine n^{2} and -3n^{2} to get -2n^{2}.

$$-2n^{2}+4n=-9n+3$$

Add 9n to both sides.

$$-2n^{2}+4n+9n=3$$

Combine 4n and 9n to get 13n.

$$-2n^{2}+13n=3$$

Divide both sides by -2.

$$\frac{-2n^{2}+13n}{-2}=\frac{3}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$n^{2}+\frac{13}{-2}n=\frac{3}{-2}$$

Divide 13 by -2.

$$n^{2}-\frac{13}{2}n=\frac{3}{-2}$$

Divide 3 by -2.

$$n^{2}-\frac{13}{2}n=-\frac{3}{2}$$

Divide -\frac{13}{2}, the coefficient of the x term, by 2 to get -\frac{13}{4}. Then add the square of -\frac{13}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}-\frac{13}{2}n+\left(-\frac{13}{4}\right)^{2}=-\frac{3}{2}+\left(-\frac{13}{4}\right)^{2}$$

Square -\frac{13}{4} by squaring both the numerator and the denominator of the fraction.

$$n^{2}-\frac{13}{2}n+\frac{169}{16}=-\frac{3}{2}+\frac{169}{16}$$

Add -\frac{3}{2} to \frac{169}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$n^{2}-\frac{13}{2}n+\frac{169}{16}=\frac{145}{16}$$

Factor n^{2}-\frac{13}{2}n+\frac{169}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n-\frac{13}{4}\right)^{2}=\frac{145}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{13}{4}\right)^{2}}=\sqrt{\frac{145}{16}}$$

Simplify.

$$n-\frac{13}{4}=\frac{\sqrt{145}}{4} n-\frac{13}{4}=-\frac{\sqrt{145}}{4}$$

Add \frac{13}{4} to both sides of the equation.

$$n=\frac{\sqrt{145}+13}{4} n=\frac{13-\sqrt{145}}{4}$$