Mastering Equations And Inequalities A Comprehensive Guide

Mathematics, the queen of sciences, often presents us with intriguing challenges in the form of equations and inequalities. Tackling these problems requires a blend of algebraic manipulation, logical reasoning, and a keen eye for detail. In this comprehensive guide, we will delve into a variety of mathematical problems, offering step-by-step solutions and insightful explanations to enhance your problem-solving prowess. We will explore quadratic equations, radical equations, and inequalities, providing a solid foundation for your mathematical journey. Let's embark on this enlightening expedition together!

1. Solving the Quadratic Equation:

rac{x^2-6x+10}{x2+8x+17} = rac{(x-3)^2}{(x+4)2}

This equation presents a fascinating challenge, combining rational expressions and quadratic forms. To solve it effectively, we need to skillfully manipulate the equation, eliminating fractions and simplifying terms. Our primary goal is to transform the equation into a more manageable form, ideally a polynomial equation that we can solve using standard techniques.

Initial Assessment

First, let's observe the structure of the equation. We have a rational expression on both sides, with quadratic expressions in both the numerators and denominators. A direct approach might involve cross-multiplication, which can lead to a quartic equation. However, before we jump into that, let's examine if there are any clever simplifications or substitutions we can make. The key here is to look for patterns and potential cancellations.

Cross-Multiplication and Expansion

Let's begin by cross-multiplying to eliminate the fractions. This gives us:

${(x^2 - 6x + 10)(x + 4)^2 = (x - 3)^2(x^2 + 8x + 17)}$

Next, we need to expand both sides. Expanding extit{$(x + 4)^2$} gives us $x^2 + 8x + 16$, and extit{$(x - 3)^2$} gives us $x^2 - 6x + 9$. Substituting these back into the equation, we get:

${(x^2 - 6x + 10)(x^2 + 8x + 16) = (x^2 - 6x + 9)(x^2 + 8x + 17)}$

Now, we need to expand the products on both sides. This is where the algebra gets a bit intensive:

Left-hand side (LHS):

${(x^2 - 6x + 10)(x^2 + 8x + 16) = x^4 + 8x^3 + 16x^2 - 6x^3 - 48x^2 - 96x + 10x^2 + 80x + 160}$

Simplifying the LHS, we have:

${x^4 + 2x^3 - 22x^2 - 16x + 160}$

Right-hand side (RHS):

${(x^2 - 6x + 9)(x^2 + 8x + 17) = x^4 + 8x^3 + 17x^2 - 6x^3 - 48x^2 - 102x + 9x^2 + 72x + 153}$

Simplifying the RHS, we have:

${x^4 + 2x^3 - 22x^2 - 30x + 153}$

Forming the Polynomial Equation

Now, we set the LHS equal to the RHS:

${x^4 + 2x^3 - 22x^2 - 16x + 160 = x^4 + 2x^3 - 22x^2 - 30x + 153}$

Notice that several terms cancel out: $x^4$, $2x^3$, and $-22x^2$. This simplifies the equation significantly:

${-16x + 160 = -30x + 153}$

Solving for x

Now, we solve for $x$:

${30x - 16x = 153 - 160}$

${14x = -7}$

{x = -rac{7}{14}}

{x = -rac{1}{2}}

Verification

It's crucial to verify our solution by substituting x = -rac{1}{2} back into the original equation. This step ensures that our solution is valid and that we haven't introduced any extraneous roots during the algebraic manipulations.

Substituting x = -rac{1}{2} into the original equation:

{rac{(-rac{1}{2})^2 - 6(-rac{1}{2}) + 10}{(-rac{1}{2})^2 + 8(-rac{1}{2}) + 17} = rac{(-rac{1}{2} - 3)^2}{(-rac{1}{2} + 4)^2}}

Simplifying:

{rac{rac{1}{4} + 3 + 10}{rac{1}{4} - 4 + 17} = rac{(-rac{7}{2})^2}{(rac{7}{2})^2}}

{rac{rac{1}{4} + 13}{rac{1}{4} + 13} = rac{rac{49}{4}}{rac{49}{4}}}

{rac{rac{53}{4}}{rac{53}{4}} = 1}

${1 = 1}$

Since the equation holds true, our solution x = -rac{1}{2} is indeed valid.

Conclusion

Therefore, the solution to the equation is x = -rac{1}{2}. This problem highlighted the importance of careful algebraic manipulation, expansion, and simplification. Always remember to verify your solutions to ensure accuracy and avoid extraneous roots.

Key Takeaways

• Cross-multiplication is a useful technique for eliminating fractions in equations.
• Expanding and simplifying expressions are crucial steps in solving algebraic equations.
• Verification of solutions is essential to ensure accuracy.

2. Solving the Radical Equation:

${

\sqrt{x+11} + 5\sqrt{2x-3} + \sqrt{x+3} + 3\sqrt{2x-3} = 9\sqrt{2} }$

Radical equations, which involve square roots and other radicals, require a strategic approach to isolate and eliminate the radical terms. This problem presents a combination of square roots, making it a bit more complex. Our goal is to isolate the radicals and then square the equation (or perform a similar operation) to remove the square roots.

Initial Assessment

Observe the equation closely. We have four radical terms involving $\sqrt{x+11}$, $\sqrt{2x-3}$, and $\sqrt{x+3}$. The presence of multiple radicals suggests that we may need to square the equation more than once to eliminate all square roots. The key is to extit{group similar terms} and isolate the most complex radicals first.

Combining Like Terms

Let's begin by combining the terms with the same radicals:

${\sqrt{x+11} + \sqrt{x+3} + 5\sqrt{2x-3} + 3\sqrt{2x-3} = 9\sqrt{2}}$

Combining the terms with $\sqrt{2x-3}$, we get:

${\sqrt{x+11} + \sqrt{x+3} + 8\sqrt{2x-3} = 9\sqrt{2}}$

Isolating Radicals

Now, let's isolate one of the radicals. We'll isolate the $8\sqrt{2x-3}$ term:

${8\sqrt{2x-3} = 9\sqrt{2} - \sqrt{x+11} - \sqrt{x+3}}$

Squaring the Equation

To eliminate the square root, we square both sides of the equation:

${(8\sqrt{2x-3})^2 = (9\sqrt{2} - \sqrt{x+11} - \sqrt{x+3})^2}$

This simplifies to:

${64(2x-3) = (9\sqrt{2} - \sqrt{x+11} - \sqrt{x+3})^2}$

Expanding the left side, we have:

${128x - 192 = (9\sqrt{2} - \sqrt{x+11} - \sqrt{x+3})^2}$

Expanding the right side is more complex. Let's use the formula $(a - b - c)^2 = a^2 + b^2 + c^2 - 2ab - 2ac + 2bc$:

${(9\sqrt{2})^2 + (\sqrt{x+11})^2 + (\sqrt{x+3})^2 - 2(9\sqrt{2})(\sqrt{x+11}) - 2(9\sqrt{2})(\sqrt{x+3}) + 2(\sqrt{x+11})(\sqrt{x+3})}$

This simplifies to:

${162 + (x+11) + (x+3) - 18\sqrt{2(x+11)} - 18\sqrt{2(x+3)} + 2\sqrt{(x+11)(x+3)}}$

Combining like terms, we have:

${162 + x + 11 + x + 3 - 18\sqrt{2x+22} - 18\sqrt{2x+6} + 2\sqrt{x^2+14x+33}}$

${176 + 2x - 18\sqrt{2x+22} - 18\sqrt{2x+6} + 2\sqrt{x^2+14x+33}}$

So, our equation now looks like:

${128x - 192 = 176 + 2x - 18\sqrt{2x+22} - 18\sqrt{2x+6} + 2\sqrt{x^2+14x+33}}$

Simplifying the Equation

Rearrange the terms to isolate the remaining radicals:

${126x - 368 = - 18\sqrt{2x+22} - 18\sqrt{2x+6} + 2\sqrt{x^2+14x+33}}$

Divide the entire equation by 2 to simplify:

${63x - 184 = -9\sqrt{2x+22} - 9\sqrt{2x+6} + \sqrt{x^2+14x+33}}$

This equation is still complex, and it might be challenging to proceed with further squaring. At this point, it's worth pausing and reconsidering our approach. It seems we've created a more complicated equation than we started with. This often happens with radical equations, and it indicates we might need a different strategy.

Alternative Approach: Substitution

Let's go back to our original equation:

${\sqrt{x+11} + \sqrt{x+3} + 8\sqrt{2x-3} = 9\sqrt{2}}$

Notice the terms $\sqrt{x+11}$ and $\sqrt{x+3}$. Let's try to manipulate these terms. We can rewrite the equation by isolating the radicals differently:

${\sqrt{x+11} + \sqrt{x+3} = 9\sqrt{2} - 8\sqrt{2x-3}}$

Now, square both sides:

${(\sqrt{x+11} + \sqrt{x+3})^2 = (9\sqrt{2} - 8\sqrt{2x-3})^2}$

Expanding the left side, we get:

${(x+11) + 2\sqrt{(x+11)(x+3)} + (x+3)}$

${2x + 14 + 2\sqrt{x^2+14x+33}}$

Expanding the right side, we get:

${(9\sqrt{2})^2 - 2(9\sqrt{2})(8\sqrt{2x-3}) + (8\sqrt{2x-3})^2}$

${162 - 144\sqrt{2(2x-3)} + 64(2x-3)}$

${162 - 144\sqrt{4x-6} + 128x - 192}$

${128x - 30 - 144\sqrt{4x-6}}$

Setting the two sides equal:

${2x + 14 + 2\sqrt{x^2+14x+33} = 128x - 30 - 144\sqrt{4x-6}}$

Isolate the remaining radicals:

${2\sqrt{x^2+14x+33} + 144\sqrt{4x-6} = 126x - 44}$

Divide by 2:

${\sqrt{x^2+14x+33} + 72\sqrt{4x-6} = 63x - 22}$

This equation is still quite complex. It seems that a straightforward algebraic manipulation might not lead to a simple solution. We might need to consider numerical methods or approximation techniques to find the solution, or there might be a clever substitution we are missing. Let's try substituting a value of x to verify our result. By observation, if we substitute $x = 3$, we find

${\sqrt{3+11} + 5\sqrt{2(3)-3} + \sqrt{3+3} + 3\sqrt{2(3)-3} = 9\sqrt{2}}$

${\sqrt{14} + 5\sqrt{3} + \sqrt{6} + 3\sqrt{3} \neq 9\sqrt{2}}$

Which does not satisfy the equation. Since this equation is complex and doesn't seem to yield a simple analytical solution, we can explore numerical methods or approximations to find the roots.

Conclusion

Solving this radical equation proved to be more challenging than initially anticipated. While we made progress in simplifying the equation, we encountered significant complexity when trying to eliminate all the radicals. This highlights the importance of recognizing when a particular approach might not be the most efficient and being open to exploring alternative strategies. In this case, a numerical method might be more appropriate.

Key Takeaways

• Isolating radicals and squaring the equation are fundamental techniques for solving radical equations.
• Sometimes, straightforward algebraic manipulation can lead to complex equations, indicating the need for alternative approaches.
• Numerical methods or approximations may be necessary for solving highly complex radical equations.

3. Solving the Inequality:

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

Inequalities are mathematical statements that compare two expressions using symbols like >, <, ≥, or ≤. Solving inequalities involves finding the range of values for the variable that satisfy the inequality. This problem presents a linear inequality involving fractions, which requires careful manipulation to isolate the variable.

Initial Assessment

Our goal is to isolate $x$ on one side of the inequality. The presence of fractions suggests that we should begin by eliminating the denominators to simplify the inequality. This can be achieved by multiplying both sides by the least common multiple (LCM) of the denominators.

Eliminating Fractions

The denominators in the inequality are 5 and 3. The least common multiple (LCM) of 5 and 3 is 15. So, we multiply both sides of the inequality by 15:

${15\left(\frac{2x+1}{5} - \frac{2-x}{3}\right) > 15(1)}$

Distribute the 15 to each term:

${15\left(\frac{2x+1}{5}\right) - 15\left(\frac{2-x}{3}\right) > 15}$

Simplify each term:

${3(2x+1) - 5(2-x) > 15}$

Expanding and Simplifying

Now, expand the expressions:

${6x + 3 - 10 + 5x > 15}$

Combine like terms:

${11x - 7 > 15}$

Isolating the Variable

Add 7 to both sides of the inequality:

${11x > 15 + 7}$

${11x > 22}$

Divide both sides by 11:

${x > \frac{22}{11}}$

${x > 2}$

Solution and Interpretation

The solution to the inequality is $x > 2$. This means that any value of $x$ greater than 2 will satisfy the original inequality. We can represent this solution graphically on a number line, with an open circle at 2 and an arrow extending to the right, indicating all values greater than 2.

Verification

To verify our solution, we can test a value of $x$ that is greater than 2 and a value that is less than or equal to 2. Let's test $x = 3$ (which should satisfy the inequality) and $x = 2$ (which should not).

For $x = 3$:

${\frac{2(3)+1}{5} - \frac{2-3}{3} > 1}$

${\frac{7}{5} - \frac{-1}{3} > 1}$

${\frac{7}{5} + \frac{1}{3} > 1}$

${\frac{21+5}{15} > 1}$

${\frac{26}{15} > 1}$

This is true, so $x = 3$ satisfies the inequality.

For $x = 2$:

${\frac{2(2)+1}{5} - \frac{2-2}{3} > 1}$

${\frac{5}{5} - \frac{0}{3} > 1}$

${1 - 0 > 1}$

${1 > 1}$

This is false, so $x = 2$ does not satisfy the inequality.

Conclusion

Therefore, the solution to the inequality is $x > 2$. This problem demonstrated the process of solving a linear inequality involving fractions, including the importance of eliminating denominators, simplifying expressions, and verifying the solution.

Key Takeaways

• Multiplying by the LCM of the denominators is an effective way to eliminate fractions in inequalities.
• Careful expansion and simplification are crucial for solving inequalities accurately.
• Verification of the solution set is essential to ensure correctness.

4. Solving the Absolute Value Equation:

${\frac{|x-1|}{x} = \frac{4}{x}}$

Absolute value equations involve expressions within absolute value symbols, which represent the distance of a number from zero. Solving these equations requires considering different cases based on the sign of the expression inside the absolute value. This problem presents an absolute value equation with rational expressions, which adds an extra layer of complexity.

Initial Assessment

The presence of the absolute value $|x-1|$ means we need to consider two cases: $x-1 \geq 0$ and $x-1 < 0$. The denominator $x$ also introduces a restriction: $x$ cannot be equal to 0, as this would make the fractions undefined. Our strategy will be to solve the equation separately for each case and then check for extraneous solutions.

Case 1: $x-1 \geq 0$ (i.e., $x \geq 1$)

If $x \geq 1$, then $|x-1| = x-1$. The equation becomes:

${\frac{x-1}{x} = \frac{4}{x}}$

Since both sides have the same denominator $x$, we can equate the numerators, provided $x \neq 0$:

${x - 1 = 4}$

${x = 5}$

Since $5 \geq 1$, this solution is valid for this case.

Case 2: $x-1 < 0$ (i.e., $x < 1$)

If $x < 1$, then $|x-1| = -(x-1) = 1-x$. The equation becomes:

${\frac{1-x}{x} = \frac{4}{x}}$

Again, we equate the numerators, provided $x \neq 0$:

${1 - x = 4}$

${-x = 3}$

${x = -3}$

Since $-3 < 1$, this solution is also valid for this case.

Combining Solutions

Our potential solutions are $x = 5$ and $x = -3$. We need to verify that these solutions are not extraneous by substituting them back into the original equation.

Verification

For $x = 5$:

${\frac{|5-1|}{5} = \frac{4}{5}}$

${\frac{|4|}{5} = \frac{4}{5}}$

${\frac{4}{5} = \frac{4}{5}}$

This is true, so $x = 5$ is a valid solution.

For $x = -3$:

${\frac{|-3-1|}{-3} = \frac{4}{-3}}$

${\frac{|-4|}{-3} = \frac{4}{-3}}$

${\frac{4}{-3} = \frac{4}{-3}}$

This is also true, so $x = -3$ is a valid solution.

Conclusion

Therefore, the solutions to the absolute value equation are $x = 5$ and $x = -3$. This problem highlighted the importance of considering different cases when solving absolute value equations and the necessity of verifying solutions to avoid extraneous roots.

Key Takeaways

• Absolute value equations require considering different cases based on the sign of the expression inside the absolute value.
• Equating numerators is possible when both sides of an equation have the same denominator (provided the denominator is not zero).
• Verification of solutions is essential to ensure accuracy and avoid extraneous roots.

Conclusion

In this comprehensive guide, we've navigated through a diverse range of mathematical equations and inequalities, from quadratic and radical equations to linear inequalities and absolute value equations. Each problem presented unique challenges and required a combination of algebraic manipulation, logical reasoning, and attention to detail. We've emphasized the importance of systematic problem-solving approaches, verification of solutions, and recognizing when alternative strategies might be necessary.

Mastering these concepts and techniques will significantly enhance your mathematical skills and problem-solving abilities. Remember, mathematics is a journey of exploration and discovery, and with practice and perseverance, you can conquer even the most intricate problems. Keep exploring, keep learning, and keep pushing your mathematical boundaries!