Rationalizing Denominators And Representing Square Roots On The Number Line

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1. Simplifying Expressions by Rationalizing the Denominator

Understanding Rationalization

Rationalizing the denominator is a technique used to eliminate radical expressions (usually square roots) from the denominator of a fraction. This process involves multiplying both the numerator and the denominator by a suitable expression, which transforms the denominator into a rational number without changing the value of the fraction. Rationalizing the denominator is not just a matter of mathematical aesthetics; it simplifies further calculations and makes it easier to compare and manipulate expressions. The core idea behind this technique is to exploit the difference of squares identity, which states that (a+b)(a−b)=a2−b2{(a + b)(a - b) = a^2 - b^2}. By multiplying the denominator by its conjugate, we can eliminate the square root, effectively rationalizing the denominator. This method ensures that the denominator becomes a rational number, making the entire expression easier to work with in subsequent calculations or comparisons.

i) Simplifying

6−426+42{\frac{6-4\sqrt{2}}{6+4\sqrt{2}}}

To simplify the expression 6−426+42{\frac{6-4\sqrt{2}}{6+4\sqrt{2}}} by rationalizing the denominator, we need to multiply both the numerator and the denominator by the conjugate of the denominator. The conjugate of 6+42{6+4\sqrt{2}} is 6−42{6-4\sqrt{2}}. This process leverages the difference of squares, eliminating the square root in the denominator. Here's a step-by-step breakdown:

  1. Identify the Conjugate: The conjugate of the denominator 6+42{6 + 4\sqrt{2}} is 6−42{6 - 4\sqrt{2}}.
  2. Multiply by the Conjugate: Multiply both the numerator and the denominator by the conjugate: 6−426+42×6−426−42{\frac{6-4\sqrt{2}}{6+4\sqrt{2}} \times \frac{6-4\sqrt{2}}{6-4\sqrt{2}}}
  3. Expand the Numerator: Multiply out the terms in the numerator: (6−42)(6−42)=6(6)+6(−42)−42(6)+(−42)(−42){(6 - 4\sqrt{2})(6 - 4\sqrt{2}) = 6(6) + 6(-4\sqrt{2}) - 4\sqrt{2}(6) + (-4\sqrt{2})(-4\sqrt{2})} =36−242−242+16(2)=36−482+32=68−482{= 36 - 24\sqrt{2} - 24\sqrt{2} + 16(2) = 36 - 48\sqrt{2} + 32 = 68 - 48\sqrt{2}}
  4. Expand the Denominator: Multiply out the terms in the denominator using the difference of squares formula, (a+b)(a−b)=a2−b2{(a+b)(a-b) = a^2 - b^2}: (6+42)(6−42)=62−(42)2=36−16(2)=36−32=4{(6 + 4\sqrt{2})(6 - 4\sqrt{2}) = 6^2 - (4\sqrt{2})^2 = 36 - 16(2) = 36 - 32 = 4}
  5. Simplify the Fraction: Now, the fraction looks like this: 68−4824{\frac{68 - 48\sqrt{2}}{4}}
  6. Divide by the Common Factor: Divide both terms in the numerator by the denominator: 684−4824=17−122{\frac{68}{4} - \frac{48\sqrt{2}}{4} = 17 - 12\sqrt{2}}

Therefore, the simplified form of 6−426+42{\frac{6-4\sqrt{2}}{6+4\sqrt{2}}} is 17−122{17 - 12\sqrt{2}}. This process not only removes the square root from the denominator but also simplifies the expression into a more manageable form. The technique of multiplying by the conjugate is a cornerstone in algebraic manipulations, particularly when dealing with irrational numbers.

ii) Simplifying

3053−35{\frac{30}{5\sqrt{3}-3\sqrt{5}}}

Simplifying the expression 3053−35{\frac{30}{5\sqrt{3}-3\sqrt{5}}} also involves rationalizing the denominator. Again, we multiply both the numerator and the denominator by the conjugate of the denominator. The conjugate of 53−35{5\sqrt{3} - 3\sqrt{5}} is 53+35{5\sqrt{3} + 3\sqrt{5}}. This approach ensures that the denominator will be rationalized by eliminating the square roots. Here's a detailed breakdown of the steps:

  1. Identify the Conjugate: The conjugate of the denominator 53−35{5\sqrt{3} - 3\sqrt{5}} is 53+35{5\sqrt{3} + 3\sqrt{5}}.
  2. Multiply by the Conjugate: Multiply both the numerator and the denominator by the conjugate: 3053−35×53+3553+35{\frac{30}{5\sqrt{3}-3\sqrt{5}} \times \frac{5\sqrt{3}+3\sqrt{5}}{5\sqrt{3}+3\sqrt{5}}}
  3. Expand the Numerator: Multiply the numerator: 30(53+35)=1503+905{30(5\sqrt{3} + 3\sqrt{5}) = 150\sqrt{3} + 90\sqrt{5}}
  4. Expand the Denominator: Multiply out the terms in the denominator using the difference of squares formula (a+b)(a−b)=a2−b2{(a+b)(a-b) = a^2 - b^2}: (53−35)(53+35)=(53)2−(35)2{(5\sqrt{3} - 3\sqrt{5})(5\sqrt{3} + 3\sqrt{5}) = (5\sqrt{3})^2 - (3\sqrt{5})^2} =25(3)−9(5)=75−45=30{= 25(3) - 9(5) = 75 - 45 = 30}
  5. Simplify the Fraction: Now, the fraction is: 1503+90530{\frac{150\sqrt{3} + 90\sqrt{5}}{30}}
  6. Divide by the Common Factor: Divide both terms in the numerator by the denominator: 150330+90530=53+35{\frac{150\sqrt{3}}{30} + \frac{90\sqrt{5}}{30} = 5\sqrt{3} + 3\sqrt{5}}

Therefore, the simplified form of 3053−35{\frac{30}{5\sqrt{3}-3\sqrt{5}}} is 53+35{5\sqrt{3} + 3\sqrt{5}}. This simplification process highlights the power of using conjugates to rationalize denominators and reduce complex expressions into simpler forms. Mastering this technique is essential for advanced algebraic manipulations.

2. Representing

5.6{\sqrt{5.6}} on the Number Line

Representing square roots on the number line is a classic geometric construction that beautifully illustrates the connection between algebra and geometry. To represent 5.6{\sqrt{5.6}} on the number line, we use a geometric method based on the properties of right triangles and circles. This construction not only provides a visual representation of irrational numbers but also enhances understanding of geometric principles. The method involves several steps, each grounded in geometric theorems and constructions, making it a valuable exercise in mathematical reasoning.

Step-by-Step Construction

  1. Draw a Line Segment: Draw a line segment AB of length 5.6 units on the number line. This line segment represents the number under the square root and forms the basis of our construction. The accuracy of this initial step is crucial for the final representation. Ensure that the length is measured precisely to reflect the value of 5.6 accurately. This line segment will be extended in the subsequent steps to facilitate the geometric construction.

  2. Extend the Line Segment: Extend the line segment AB to point C such that BC is 1 unit. The additional 1 unit is a critical component in constructing the semicircle that will help us find the square root. The length AC is now 5.6 + 1 = 6.6 units. This extension provides the diameter of the semicircle, which is essential for the geometric proof underlying the construction.

  3. Find the Midpoint: Find the midpoint O of AC. This can be done by bisecting the line segment AC. The midpoint will serve as the center of the semicircle. Bisecting the line segment accurately is important for ensuring the semicircle is drawn correctly, which in turn affects the precision of the final result. The midpoint is found by using a compass to draw arcs from points A and C and then connecting the intersection points of the arcs.

  4. Draw a Semicircle: With O as the center and OA (or OC) as the radius, draw a semicircle. This semicircle is the cornerstone of the geometric construction. The radius of the semicircle is half the length of AC, which is 6.62=3.3{\frac{6.6}{2} = 3.3} units. The semicircle's properties, particularly the relationship between its radius, diameter, and inscribed angles, are crucial to the method's success.

  5. Draw a Perpendicular: At point B, draw a line BD perpendicular to AC. This line intersects the semicircle at point D. The perpendicular line at B creates a right-angled triangle within the semicircle, which is essential for applying the geometric mean theorem. The perpendicular should be drawn accurately using a compass and straightedge or a set square to ensure the angle is exactly 90 degrees.

  6. Measure BD: The length of BD represents 5.6{\sqrt{5.6}}. Measure the length of BD using a ruler. The length BD is the geometric mean of AB and BC, which is the square root of their product. This is a direct application of the geometric mean theorem, which is the mathematical basis for this construction. The measurement of BD should be done carefully to get an accurate representation of 5.6{\sqrt{5.6}}.

Mathematical Justification

The geometric construction is based on the geometric mean theorem, which states that in a right-angled triangle, the altitude to the hypotenuse divides the hypotenuse into two segments, and the length of the altitude is the geometric mean of these segments. In our construction, triangle ADC is a right-angled triangle because it is inscribed in a semicircle. BD is the altitude to the hypotenuse AC. According to the geometric mean theorem: BD2=AB×BC{BD^2 = AB \times BC} Since AB = 5.6 units and BC = 1 unit: BD2=5.6×1=5.6{BD^2 = 5.6 \times 1 = 5.6} Taking the square root of both sides: BD=5.6{BD = \sqrt{5.6}} Thus, the length of BD represents 5.6{\sqrt{5.6}} units on the number line. This method provides a visually intuitive and geometrically sound way to represent square roots of non-perfect square numbers. The use of the semicircle and the geometric mean theorem showcases the elegant interplay between geometric shapes and algebraic concepts.

Placing on the Number Line

To place 5.6{\sqrt{5.6}} accurately on the number line, mark a point E on the number line such that BE = BD. This transfers the length of BD, which represents 5.6{\sqrt{5.6}}, onto the number line, providing a precise location for the value. Using a compass, place one end at B and the other at D, and then draw an arc that intersects the number line. The point of intersection, E, represents 5.6{\sqrt{5.6}} on the number line. This step ensures that the numerical value of 5.6{\sqrt{5.6}} is clearly marked and can be easily located and compared with other values on the number line.

Conclusion

In conclusion, rationalizing denominators and representing square roots on the number line are essential skills in mathematics. The techniques discussed in this article not only simplify mathematical expressions but also provide a deeper understanding of the relationship between numbers and geometry. Mastering these concepts will undoubtedly enhance your mathematical problem-solving abilities and provide a solid foundation for more advanced topics. Rationalizing the denominator allows us to work with fractions in a more manageable form, while representing square roots on the number line provides a visual and intuitive understanding of irrational numbers. Both skills are crucial for success in mathematics and its applications.