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 $x$

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

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

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

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

Problem 2: Given

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

Find $(x-1)^{10}$

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

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:

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:

$x-1=\sqrt{3}$

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

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

Finally:

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

$(x-1)^{10}=243$

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

Problem 3: Solve for $x$

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

Solution

[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: $\{-\sqrt{3}\; , \sqrt{3}\; and\; 4\}$

Problem 4: Solve for $x$

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

Solution

$15x^{3}+38x^{2}+17x+2=0$ “

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

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

Problem 5: Solve for $x$

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

Solution

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

Problem 6: Solve for $x$

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

Solution

$6x^{3}-29x^{2}-45x+18=0$ “

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

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

Problem 7: Solve for $x$

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

Solution

$3x^{3}+3x^{2}-4x+4=0$ “

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

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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Selected Solutions

Selected Solutions

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