Selected Solutions

Selected Solutions

The following is a sample of many of the exercises we plan to solve, to show how some of the seemingly difficult problems can be easily solved.

Problem 1: Solve for x

2^{x} \cdot 6^{x-2}=5^{2x}\cdot 7^{1-x}

 

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2^{x}\cdot 6^{x-2}=5^{2x}\cdot 7^{1-x}
x\ln 2+ (x-2)\ln 6=2x\ln 5+ (1-x)\ln 7
x \ln 2+ x\ln 6- 2\ln 6= x(2\ln 5)+\ln 7-x\ln7
x\left(\ln 2+\ln 6- 2\ln 5+ \ln 7 \right)=2\ln 6+ \ln 7

Now isolating x:

x=\frac{\ln 36+ \ln 7}{\ln 2+ \ln 6- \ln 25 + \ln 7}

x= \frac{\ln (36\times 7)}{\ln (\frac{2\times 6 \times 7}{25})}

x=\frac{\ln 252}{\ln (\frac{84}{25})}

Finally:

Answer: \log_{\frac{84}{25}} 252

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Problem 2: Given

x=\frac{\sqrt{3}+\sqrt{5}+\sqrt{6}+3+\sqrt{15}+\sqrt{18}}{\sqrt{5}+\sqrt{3}(1+\sqrt{2})}

Find (x-1)^{10}

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x=\frac{\sqrt{3}+\sqrt{5}+\sqrt{6}+3+\sqrt{15}+\sqrt{18}}{\sqrt{5}+\sqrt{3}(1+\sqrt{2})}

Let’s find x-1

x-1=\frac{\sqrt{3}+\sqrt{5}+\sqrt{6}+3+\sqrt{15}+\sqrt{18}}{\sqrt{5}+\sqrt{3}(1+\sqrt{2})}-1

x-1=\frac{\sqrt{3}+\sqrt{5}+\sqrt{6}+3+\sqrt{15}+\sqrt{18}-(\sqrt{5}+\sqrt{3}(1+\sqrt{2}))}{\sqrt{5}+\sqrt{3}(1+\sqrt{2})}

x-1=\frac{\sqrt{3}+\sqrt{5}+\sqrt{3}\cdot \sqrt{2}+\sqrt{3}\cdot \sqrt{3}+\sqrt{5}\cdot \sqrt{3}+\sqrt{3}\cdot \sqrt{3}\cdot \sqrt{2}-\sqrt{5}-\sqrt{3}-\sqrt{3}\cdot\sqrt{2}}{\sqrt{5}+\sqrt{3}+\sqrt{3}\cdot \sqrt{2}}

Let’s simplify:

x-1=\frac{\sqrt{3}+\sqrt{5}+\sqrt{3}\cdot \sqrt{2}+\sqrt{3}\cdot \sqrt{3}+\sqrt{5}\cdot \sqrt{3}+\sqrt{3}\cdot \sqrt{3}\cdot \sqrt{2}-\sqrt{5}-\sqrt{3}-\sqrt{3}\cdot\sqrt{2}}{\sqrt{5}+\sqrt{3}+\sqrt{3}\cdot \sqrt{2}}

We get

x-1=\frac{\sqrt{3}\cdot \sqrt{3}+\sqrt{5}\cdot \sqrt{3}+\sqrt{3}\cdot \sqrt{3}\cdot \sqrt{2}}{\sqrt{5}+\sqrt{3}+\sqrt{3}\cdot \sqrt{2}}

x-1=\frac{\sqrt{3}( \sqrt{3}+\sqrt{5}+\sqrt{3}\cdot \sqrt{2})}{\sqrt{5}+\sqrt{3}+\sqrt{3}\cdot \sqrt{2}}

Let’s simplify:

x-1=\frac{\sqrt{3}( \sqrt{3}+\sqrt{5}+\sqrt{3}\cdot \sqrt{2})}{(\sqrt{5}+\sqrt{3}+\sqrt{3}\cdot \sqrt{2})}

 

We get:

x-1=\sqrt{3}

(x-1)^{10}=(\sqrt{3})^{10}

(x-1)^{10}=((\sqrt{3})^{2})^{5}

(x-1)^{10}=((\sqrt{3})^{2})^{5}

Finally:

(x-1)^{10}=3^{5}

(x-1)^{10}=243

Answer: (x-1)^{10}=243

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Problem 3: Solve for x

x^{3}-4x^{2}-3x+12=0

 Solution

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[item title=”Click here to see the solution of: {x}^{3}-4x^{2}-3x+12=0“]

{x}^{3}-4x^{2}-3x+12=0

We can see the factors:

{x}^{3}-4x^{2}-3x+12=x^{2}(x-4)-3(x-4)=(x^{2}-3)(x-4)=(x-\sqrt{3})(x+\sqrt{3})(x-4)

Now we can say:

(x-\sqrt{3})(x+\sqrt{3})(x-4)=0

Any factor can be a zero:

(x-\sqrt{3})=0 \Rightarrow x_{1}=\sqrt{3}

(x+\sqrt{3})=0 \Rightarrow x_{2}=-\sqrt{3}

(x-4)=0 \Rightarrow x_{3}=4

Roots: \{-\sqrt{3}\; , \sqrt{3}\; and\; 4\}

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Problem 4: Solve for x

15x^{3}+38x^{2}+17x+2=0

 Solution

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[item title=”Click here to see the solution of: 15x^{3}+38x^{2}+17x+2=0“]

 15x^{3}+38x^{2}+17x+2=0

Looking At factors and ratio of 2 and 15 we can see that -2 is a root.

When we divide:

\frac{15x^{3}+38x^{2}+17x+2}{x+2}=15x^{2}+8x+1

We now solve the quadratic:

15x^{2}+8x+1=0

\Delta=64-60=4

\sqrt{\Delta}=2

x_{2}=\frac{-8+2}{2\times 15}=-\frac{1}{5}

x_{3}=\frac{-8-2}{2\times 15}=-\frac{1}{3}

 

Using our general method without the factors of 2 and 15:

We have:

p=-1.005925926

q=0.380620027

\Delta=-0.001481481

Case 

\Delta<0

\cos 3\theta=-0.9801544601

\frac{\theta}{3}=0.9806784972

We get the same x values.

 

Roots: \{-2\; ,-\frac{1}{5}\; and \; -\frac{1}{3}\}

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Problem 5: Solve for x

18x^{4}+15x^{3}-34x^{2}+15x-2=0

 Solution

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[item title=”Click here to see the solution of: 18x^{4}+15x^{3}-34x^{2}+15x-2=0“]

18x^{4}+15x^{3}-34x^{2}+15x-2=0

By inspecting the factors of -2 and 18 and their ratios, we discover that -2 and \frac{1}{2} are roots.

Facroring these 2 we get:

(x+2)(x-\frac{1}{2})=\frac{2x^{2}+3x-2}{2}

Since these are roots, let’s drop the denominator.

\frac{18x^{4}+15x^{3}-34x^{2}+15x-2}{2x^{2}+3x-2}=9x^{2}-6x+1

Solving:9x^{2}-6x+1=0

Double root since \Delta=0

x_{3}=x_{4}=\frac{1}{3}

Roots: \{-2\; ,\frac{1}{3}\; , \frac{1}{3}\; and \; \frac{1}{2}\}

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Problem 6: Solve for x

6x^{3}-29x^{2}-45x+18=0

 Solution

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[item title=”Click here to see the solution of: 6x^{3}-29x^{2}-45x+18=0“]

6x^{3}-29x^{2}-45x+18=0

We can see that 6 is a root:

\frac{6x^{3}-29x^{2}-45x+18}{x-6}=6x^{2}+7x-3

Factoring the quadratic:

6x^{2}+7x-3=6x^{2}+9x-2x-3=3x(2x+3)-1(2x+3)=(3x-1)(2x+3)

Finally we can say:

6x^{3}-29x^{2}-45x+18=(x-6)(3x-1)(2x+3)

And to solve, any facor can be 0:

x-6=0 \Rightarrow x_{1}=6

3x-1=0 \Rightarrow x_{2}=\frac{1}{3}

2x+3=0 \Rightarrow x_{3}=-\frac{3}{2}

Finally:

Roots: \{-\frac{3}{2}\; ,\frac{1}{3}\; , \frac{1}{3}\; and \; 6\}

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Problem 7: Solve for x

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

 Solution

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[item title=”Click here to see the solution of: 3x^{3}+3x^{2}-4x+4=0“]

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

We can see that -2 is a root.

When we divide, we get:

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

Solving for the quadratic, we get complex roots:

x_{2}=\frac{1}{2}+i\frac{\sqrt{15}}{6}

x_{3}=\frac{1}{2}-i\frac{\sqrt{15}}{6}

 

Verification using the general method:

p=-1.666666667

q=1.851851852

\Delta=0.6858710562

\sin \theta=0.4472135955

\frac{\theta}{2}=0.2318238045

\tan \phi=0.6180339887

2\phi=1.107148718

The roots are the same as above.

Roots: \{-2\; ,\frac{1}{2}-i\frac{\sqrt{15}}{6}\; and \; \frac{1}{2}+i\frac{\sqrt{15}}{6}\}

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