Solve 6sqrt{N}+5sqrt{1/N}=13

$$6 \sqrt { N } + 5 \sqrt { \frac { 1 } { N } } = 13$$

Solve for N

$$N=\frac{1}{4}=0.25$$

$$N = \frac{25}{9} = 2\frac{7}{9} \approx 2.777777778$$

Solution Steps

Subtract 5\sqrt{\frac{1}{N}} from both sides of the equation.

$$6\sqrt{N}=13-5\sqrt{\frac{1}{N}}$$

Square both sides of the equation.

$$\left(6\sqrt{N}\right)^{2}=\left(13-5\sqrt{\frac{1}{N}}\right)^{2}$$

Expand \left(6\sqrt{N}\right)^{2}.

$$6^{2}\left(\sqrt{N}\right)^{2}=\left(13-5\sqrt{\frac{1}{N}}\right)^{2}$$

Calculate 6 to the power of 2 and get 36.

$$36\left(\sqrt{N}\right)^{2}=\left(13-5\sqrt{\frac{1}{N}}\right)^{2}$$

Calculate \sqrt{N} to the power of 2 and get N.

$$36N=\left(13-5\sqrt{\frac{1}{N}}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(13-5\sqrt{\frac{1}{N}}\right)^{2}.

$$36N=169-130\sqrt{\frac{1}{N}}+25\left(\sqrt{\frac{1}{N}}\right)^{2}$$

Calculate \sqrt{\frac{1}{N}} to the power of 2 and get \frac{1}{N}.

$$36N=169-130\sqrt{\frac{1}{N}}+25\times \frac{1}{N}$$

Express 25\times \frac{1}{N} as a single fraction.

$$36N=169-130\sqrt{\frac{1}{N}}+\frac{25}{N}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 169 times \frac{N}{N}.

$$36N=\frac{169N}{N}-130\sqrt{\frac{1}{N}}+\frac{25}{N}$$

Since \frac{169N}{N} and \frac{25}{N} have the same denominator, add them by adding their numerators.

$$36N=\frac{169N+25}{N}-130\sqrt{\frac{1}{N}}$$

Multiply both sides of the equation by N.

$$36NN=169N+25-130\sqrt{\frac{1}{N}}N$$

Reorder the terms.

$$36NN=169N-130\sqrt{\frac{1}{N}}N+25$$

Subtract 169N+25 from both sides of the equation.

$$36NN-\left(169N+25\right)=-130\sqrt{\frac{1}{N}}N$$

Multiply N and N to get N^{2}.

$$36N^{2}-\left(169N+25\right)=-130\sqrt{\frac{1}{N}}N$$

To find the opposite of 169N+25, find the opposite of each term.

$$36N^{2}-169N-25=-130\sqrt{\frac{1}{N}}N$$

Square both sides of the equation.

$$\left(36N^{2}-169N-25\right)^{2}=\left(-130\sqrt{\frac{1}{N}}N\right)^{2}$$

Square 36N^{2}-169N-25.

$$1296N^{4}-12168N^{3}+26761N^{2}+8450N+625=\left(-130\sqrt{\frac{1}{N}}N\right)^{2}$$

Expand \left(-130\sqrt{\frac{1}{N}}N\right)^{2}.

$$1296N^{4}-12168N^{3}+26761N^{2}+8450N+625=\left(-130\right)^{2}\left(\sqrt{\frac{1}{N}}\right)^{2}N^{2}$$

Calculate -130 to the power of 2 and get 16900.

$$1296N^{4}-12168N^{3}+26761N^{2}+8450N+625=16900\left(\sqrt{\frac{1}{N}}\right)^{2}N^{2}$$

Calculate \sqrt{\frac{1}{N}} to the power of 2 and get \frac{1}{N}.

$$1296N^{4}-12168N^{3}+26761N^{2}+8450N+625=16900\times \frac{1}{N}N^{2}$$

Express 16900\times \frac{1}{N} as a single fraction.

$$1296N^{4}-12168N^{3}+26761N^{2}+8450N+625=\frac{16900}{N}N^{2}$$

Express \frac{16900}{N}N^{2} as a single fraction.

$$1296N^{4}-12168N^{3}+26761N^{2}+8450N+625=\frac{16900N^{2}}{N}$$

Cancel out N in both numerator and denominator.

$$1296N^{4}-12168N^{3}+26761N^{2}+8450N+625=16900N$$

Subtract 16900N from both sides.

$$1296N^{4}-12168N^{3}+26761N^{2}+8450N+625-16900N=0$$

Combine 8450N and -16900N to get -8450N.

$$1296N^{4}-12168N^{3}+26761N^{2}-8450N+625=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 625 and q divides the leading coefficient 1296. List all candidates \frac{p}{q}.

$$±\frac{625}{1296},±\frac{625}{648},±\frac{625}{432},±\frac{625}{324},±\frac{625}{216},±\frac{625}{162},±\frac{625}{144},±\frac{625}{108},±\frac{625}{81},±\frac{625}{72},±\frac{625}{54},±\frac{625}{48},±\frac{625}{36},±\frac{625}{27},±\frac{625}{24},±\frac{625}{18},±\frac{625}{16},±\frac{625}{12},±\frac{625}{9},±\frac{625}{8},±\frac{625}{6},±\frac{625}{4},±\frac{625}{3},±\frac{625}{2},±625,±\frac{125}{1296},±\frac{125}{648},±\frac{125}{432},±\frac{125}{324},±\frac{125}{216},±\frac{125}{162},±\frac{125}{144},±\frac{125}{108},±\frac{125}{81},±\frac{125}{72},±\frac{125}{54},±\frac{125}{48},±\frac{125}{36},±\frac{125}{27},±\frac{125}{24},±\frac{125}{18},±\frac{125}{16},±\frac{125}{12},±\frac{125}{9},±\frac{125}{8},±\frac{125}{6},±\frac{125}{4},±\frac{125}{3},±\frac{125}{2},±125,±\frac{25}{1296},±\frac{25}{648},±\frac{25}{432},±\frac{25}{324},±\frac{25}{216},±\frac{25}{162},±\frac{25}{144},±\frac{25}{108},±\frac{25}{81},±\frac{25}{72},±\frac{25}{54},±\frac{25}{48},±\frac{25}{36},±\frac{25}{27},±\frac{25}{24},±\frac{25}{18},±\frac{25}{16},±\frac{25}{12},±\frac{25}{9},±\frac{25}{8},±\frac{25}{6},±\frac{25}{4},±\frac{25}{3},±\frac{25}{2},±25,±\frac{5}{1296},±\frac{5}{648},±\frac{5}{432},±\frac{5}{324},±\frac{5}{216},±\frac{5}{162},±\frac{5}{144},±\frac{5}{108},±\frac{5}{81},±\frac{5}{72},±\frac{5}{54},±\frac{5}{48},±\frac{5}{36},±\frac{5}{27},±\frac{5}{24},±\frac{5}{18},±\frac{5}{16},±\frac{5}{12},±\frac{5}{9},±\frac{5}{8},±\frac{5}{6},±\frac{5}{4},±\frac{5}{3},±\frac{5}{2},±5,±\frac{1}{1296},±\frac{1}{648},±\frac{1}{432},±\frac{1}{324},±\frac{1}{216},±\frac{1}{162},±\frac{1}{144},±\frac{1}{108},±\frac{1}{81},±\frac{1}{72},±\frac{1}{54},±\frac{1}{48},±\frac{1}{36},±\frac{1}{27},±\frac{1}{24},±\frac{1}{18},±\frac{1}{16},±\frac{1}{12},±\frac{1}{9},±\frac{1}{8},±\frac{1}{6},±\frac{1}{4},±\frac{1}{3},±\frac{1}{2},±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$N=\frac{1}{9}$$

By Factor theorem, N-k is a factor of the polynomial for each root k. Divide 1296N^{4}-12168N^{3}+26761N^{2}-8450N+625 by 9\left(N-\frac{1}{9}\right)=9N-1 to get 144N^{3}-1336N^{2}+2825N-625. Solve the equation where the result equals to 0.

$$144N^{3}-1336N^{2}+2825N-625=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -625 and q divides the leading coefficient 144. List all candidates \frac{p}{q}.

$$±\frac{625}{144},±\frac{625}{72},±\frac{625}{48},±\frac{625}{36},±\frac{625}{24},±\frac{625}{18},±\frac{625}{16},±\frac{625}{12},±\frac{625}{9},±\frac{625}{8},±\frac{625}{6},±\frac{625}{4},±\frac{625}{3},±\frac{625}{2},±625,±\frac{125}{144},±\frac{125}{72},±\frac{125}{48},±\frac{125}{36},±\frac{125}{24},±\frac{125}{18},±\frac{125}{16},±\frac{125}{12},±\frac{125}{9},±\frac{125}{8},±\frac{125}{6},±\frac{125}{4},±\frac{125}{3},±\frac{125}{2},±125,±\frac{25}{144},±\frac{25}{72},±\frac{25}{48},±\frac{25}{36},±\frac{25}{24},±\frac{25}{18},±\frac{25}{16},±\frac{25}{12},±\frac{25}{9},±\frac{25}{8},±\frac{25}{6},±\frac{25}{4},±\frac{25}{3},±\frac{25}{2},±25,±\frac{5}{144},±\frac{5}{72},±\frac{5}{48},±\frac{5}{36},±\frac{5}{24},±\frac{5}{18},±\frac{5}{16},±\frac{5}{12},±\frac{5}{9},±\frac{5}{8},±\frac{5}{6},±\frac{5}{4},±\frac{5}{3},±\frac{5}{2},±5,±\frac{1}{144},±\frac{1}{72},±\frac{1}{48},±\frac{1}{36},±\frac{1}{24},±\frac{1}{18},±\frac{1}{16},±\frac{1}{12},±\frac{1}{9},±\frac{1}{8},±\frac{1}{6},±\frac{1}{4},±\frac{1}{3},±\frac{1}{2},±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$N=\frac{1}{4}$$

By Factor theorem, N-k is a factor of the polynomial for each root k. Divide 144N^{3}-1336N^{2}+2825N-625 by 4\left(N-\frac{1}{4}\right)=4N-1 to get 36N^{2}-325N+625. Solve the equation where the result equals to 0.

$$36N^{2}-325N+625=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 36 for a, -325 for b, and 625 for c in the quadratic formula.

$$N=\frac{-\left(-325\right)±\sqrt{\left(-325\right)^{2}-4\times 36\times 625}}{2\times 36}$$

Do the calculations.

$$N=\frac{325±125}{72}$$

Solve the equation 36N^{2}-325N+625=0 when ± is plus and when ± is minus.

$$N=\frac{25}{9} N=\frac{25}{4}$$

List all found solutions.

$$N=\frac{1}{9} N=\frac{1}{4} N=\frac{25}{9} N=\frac{25}{4}$$

Substitute \frac{1}{9} for N in the equation 6\sqrt{N}+5\sqrt{\frac{1}{N}}=13.

$$6\sqrt{\frac{1}{9}}+5\sqrt{\frac{1}{\frac{1}{9}}}=13$$

Simplify. The value N=\frac{1}{9} does not satisfy the equation.

$$17=13$$

Substitute \frac{1}{4} for N in the equation 6\sqrt{N}+5\sqrt{\frac{1}{N}}=13.

$$6\sqrt{\frac{1}{4}}+5\sqrt{\frac{1}{\frac{1}{4}}}=13$$

Simplify. The value N=\frac{1}{4} satisfies the equation.

$$13=13$$

Substitute \frac{25}{9} for N in the equation 6\sqrt{N}+5\sqrt{\frac{1}{N}}=13.

$$6\sqrt{\frac{25}{9}}+5\sqrt{\frac{1}{\frac{25}{9}}}=13$$

Simplify. The value N=\frac{25}{9} satisfies the equation.

$$13=13$$

Substitute \frac{25}{4} for N in the equation 6\sqrt{N}+5\sqrt{\frac{1}{N}}=13.

$$6\sqrt{\frac{25}{4}}+5\sqrt{\frac{1}{\frac{25}{4}}}=13$$

Simplify. The value N=\frac{25}{4} does not satisfy the equation.

$$17=13$$

Substitute \frac{1}{4} for N in the equation 6\sqrt{N}+5\sqrt{\frac{1}{N}}=13.

$$6\sqrt{\frac{1}{4}}+5\sqrt{\frac{1}{\frac{1}{4}}}=13$$

Simplify. The value N=\frac{1}{4} satisfies the equation.

$$13=13$$

Substitute \frac{25}{9} for N in the equation 6\sqrt{N}+5\sqrt{\frac{1}{N}}=13.

$$6\sqrt{\frac{25}{9}}+5\sqrt{\frac{1}{\frac{25}{9}}}=13$$

Simplify. The value N=\frac{25}{9} satisfies the equation.

$$13=13$$

List all solutions of 6\sqrt{N}=13-5\sqrt{\frac{1}{N}}.

$$N=\frac{1}{4} N=\frac{25}{9}$$