By the end of this chapter, you will be able to:
Mastering these skills will help you solve practical engineering problems efficiently and with confidence in your trade.
The ability to expand expressions using the binomial theorem is essential for quantity surveyors when approximating roots of numbers, particularly in cost estimations and measurements where exact values are not readily available. This chapter focuses on applying the binomial theorem to calculate roots such as square roots and cube roots, which frequently occur in construction material calculations and structural design assessments. Mastery of this technique enables efficient and accurate approximations without reliance on calculators, saving time on site and during preliminary cost analysis.
The binomial theorem provides a systematic way to expand expressions of the form \((1 + x)^n\), where \(n\) can be any real number, including fractions and negatives. This property allows us to express roots as binomial expansions, which can be truncated to approximate the root value. For quantity surveyors, this is particularly useful when dealing with measurements or unit rates that involve roots, such as calculating areas from dimensions or adjusting quantities.
The binomial theorem for any real exponent \(n\) states:
$$ (1 + x)^n = 1 + nx + \frac{n(n - 1)}{2!} x^2 + \frac{n(n - 1)(n - 2)}{3!} x^3 + \cdots $$
where \(|x| < 1\) for convergence. To find roots such as \(\sqrt{a}\), rewrite \(a\) in the form \(a = b(1 + x)\) where \(b\) is a perfect square close to \(a\), and \(x\) is a small fraction. Then,
$$ \sqrt{a} = \sqrt{b(1 + x)} = \sqrt{b} (1 + x)^{\frac{1}{2}} $$
Expanding \((1 + x)^{\frac{1}{2}}\) using the binomial theorem gives an approximate value for \(\sqrt{a}\).
Example 1: Approximate \(\sqrt{10}\) using the binomial theorem up to the \(x^2\) term.
Given: \(a = 10\), choose \(b = 9\) (since \(9\) is a perfect square near \(10\)).
Calculate \(x = \frac{10 - 9}{9} = \frac{1}{9} \approx 0.1111\).
Formula:
$$ \sqrt{10} = \sqrt{9(1 + 0.1111)} = 3(1 + 0.1111)^{\frac{1}{2}} = 3 [1 + \frac{1}{2}(0.1111) - \frac{1}{8}(0.1111)^2 ] $$
Substitution:
$$ = 3 [1 + 0.05555 - 0.00154 ] $$
Arithmetic:
$$ = 3 (1.05399) = 3.16197 $$
Answer: \(\sqrt{10} \approx 3.162\)
Example 2: Approximate \(\sqrt[3]{28}\) using the binomial theorem up to the \(x^2\) term.
Given: \(a = 28\), choose \(b = 27\) (perfect cube near 28).
Calculate \(x = \frac{28 - 27}{27} = \frac{1}{27} \approx 0.03704\).
Formula:
$$ \sqrt[3]{28} = \sqrt[3]{27(1 + 0.03704)} = 3 (1 + 0.03704)^{\frac{1}{3}} = 3 [1 + \frac{1}{3}(0.03704) - \frac{1}{9}(0.03704)^2 ] $$
Substitution:
$$ = 3 [1 + 0.01235 - 0.00015 ] $$
Arithmetic:
$$ = 3 (1.01220) = 3.0366 $$
Answer: \(\sqrt[3]{28} \approx 3.037\)
Example 3: Approximate \(\sqrt{50}\) to three terms in the binomial expansion.
Given: \(a = 50\), choose \(b = 49\).
Calculate \(x = \frac{50 - 49}{49} = \frac{1}{49} \approx 0.02041\).
Formula:
$$ \sqrt{50} = 7 (1 + 0.02041)^{\frac{1}{2}} = 7 [1 + \frac{1}{2}(0.02041) - \frac{1}{8}(0.02041)^2 + \frac{1}{16}(0.02041)^3 ] $$
Substitution:
$$ = 7 [1 + 0.01020 - 0.00005 + 0.000001 ] $$
Arithmetic:
$$ = 7 (1.01015) = 7.07105 $$
Answer: \(\sqrt{50} \approx 7.071\)
Example 4: Approximate \(\sqrt[3]{65}\) using binomial expansion up to \(x^3\).
Given: \(a = 65\), choose \(b = 64\) (perfect cube).
Calculate \(x = \frac{65 - 64}{64} = \frac{1}{64} = 0.015625\).
Formula:
$$ \sqrt[3]{65} = 4 (1 + 0.015625)^{\frac{1}{3}} = 4 [1 + \frac{1}{3}(0.015625) - \frac{1}{9}(0.015625)^2 + \frac{5}{81}(0.015625)^3 ] $$
Substitution:
$$ = 4 [1 + 0.005208 - 0.000027 + 0.000002 ] $$
Arithmetic:
$$ = 4 (1.005183) = 4.02073 $$
Answer: \(\sqrt[3]{65} \approx 4.021\)
Example 5: Approximate \(\sqrt{18}\) using binomial expansion to the second order term.
Given: \(a = 18\), choose \(b = 16\).
Calculate \(x = \frac{18 - 16}{16} = \frac{2}{16} = 0.125\).
Formula:
$$ \sqrt{18} = 4 (1 + 0.125)^{\frac{1}{2}} = 4 [1 + \frac{1}{2}(0.125) - \frac{1}{8}(0.125)^2 ] $$
Substitution:
$$ = 4 [1 + 0.0625 - 0.00195 ] $$
Arithmetic:
$$ = 4 (1.06055) = 4.2422 $$
Answer: \(\sqrt{18} \approx 4.242\)
In quantity surveying, precise measurements often require calculating roots for areas, volumes, or material quantities when exact values are unavailable. The binomial expansion method allows quantity surveyors to estimate these roots quickly and with reasonable accuracy during site measurements or preliminary cost planning. Using nearby perfect powers simplifies the expansion and makes the process practical for manual calculations.
Example 1: Estimate the side length of a square slab with area 102.4 m² using binomial expansion.
Given: Area \(A = 102.4\) m², side length \(s = \sqrt{102.4}\).
Choose \(b = 100\), \(x = \frac{102.4 - 100}{100} = 0.024\).
Formula:
$$ s = 10 (1 + 0.024)^{\frac{1}{2}} = 10 [1 + \frac{1}{2}(0.024) - \frac{1}{8}(0.024)^2 ] $$
Substitution:
$$ = 10 (1 + 0.012 - 0.000072) $$
Arithmetic:
$$ = 10 (1.011928) = 10.11928 $$
Answer: Side length \(\approx 10.12\) m
Example 2: Approximate the cube root of 1331.5 m³ to find the side of a cubic concrete block.
Given: Volume \(V = 1331.5\) m³, side length \(s = \sqrt[3]{1331.5}\).
Choose \(b = 1331\) (since \(11^3 = 1331\)).
Calculate \(x = \frac{1331.5 - 1331}{1331} = 0.000375\).
Formula:
$$ s = 11 (1 + 0.000375)^{\frac{1}{3}} = 11 [1 + \frac{1}{3}(0.000375) - \frac{1}{9}(0.000375)^2 ] $$
Substitution:
$$ = 11 (1 + 0.000125 - 0.0000000156) $$
Arithmetic:
$$ = 11 (1.0001249844) = 11.00137 $$
Answer: Side length \(\approx 11.0014\) m
Example 3: Estimate the square root of 45.8 m² for a rectangular floor dimension calculation.
Given: \(a = 45.8\).
Choose \(b = 49\), \(x = \frac{45.8 - 49}{49} = -0.0653\).
Formula:
$$ \sqrt{45.8} = 7 (1 - 0.0653)^{\frac{1}{2}} = 7 [1 + \frac{1}{2}(-0.0653) - \frac{1}{8}(-0.0653)^2 ] $$
Substitution:
$$ = 7 (1 - 0.03265 - 0.00053) $$
Arithmetic:
$$ = 7 (0.96682) = 6.7677 $$
Answer: \(\sqrt{45.8} \approx 6.768\) m
Example 4: Approximate the cube root of 1728.5 m³ for a storage tank volume measurement.
Given: \(a = 1728.5\), \(b = 1728\) (12³).
Calculate \(x = \frac{1728.5 - 1728}{1728} = 0.000289\).
Formula:
$$ \sqrt[3]{1728.5} = 12 (1 + 0.000289)^{\frac{1}{3}} = 12 [1 + \frac{1}{3}(0.000289) - \frac{1}{9}(0.000289)^2 ] $$
Substitution:
$$ = 12 (1 + 0.0000963 - 0.0000000093) $$
Arithmetic:
$$ = 12 (1.0000963) = 12.00116 $$
Answer: \(\approx 12.0012\) m
Example 5: Calculate the approximate square root of 24.5 m² for a small floor area measurement.
Given: \(a = 24.5\), \(b = 25\).
Calculate \(x = \frac{24.5 - 25}{25} = -0.02\).
Formula:
$$ \sqrt{24.5} = 5 (1 - 0.02)^{\frac{1}{2}} = 5 [1 + \frac{1}{2}(-0.02) - \frac{1}{8}(-0.02)^2 ] $$
Substitution:
$$ = 5 (1 - 0.01 - 0.00005) $$
Arithmetic:
$$ = 5 (0.98995) = 4.94975 $$
Answer: \(\sqrt{24.5} \approx 4.950\) m
When using binomial expansion to approximate roots, the value of \(x\) must be small enough to ensure the series converges rapidly and the approximation is accurate. For quantity surveyors-truncating after two or three terms usually provides sufficient precision for practical purposes. Understanding the error margin and convergence behavior is important for deciding how many terms to include in the expansion.
Example 1: Determine the error in approximating \(\sqrt{10}\) using two terms versus three terms.
Given: \(a = 10\), \(b = 9\), \(x = \frac{1}{9}\).
Two-term approximation:
$$ 3 (1 + \frac{1}{2} \times \frac{1}{9} ) = 3 \times 1.0556 = 3.1667 $$
Three-term approximation (from previous example):
$$ 3 \times 1.05399 = 3.16197 $$
Actual value:
$$ \sqrt{10} = 3.1623 $$
Errors:
Two-term error:
$$ |3.1667 - 3.1623| = 0.0044 $$
Three-term error:
$$ |3.16197 - 3.1623| = 0.00033 $$
Answer: Three-term expansion reduces error by over 10 times.
Example 2: Evaluate the convergence for approximating \(\sqrt[3]{28}\) with two and three terms.
Given: \(a=28\), \(b=27\), \(x=0.03704\).
Two-term:
$$ 3 \times (1 + \frac{1}{3} \times 0.03704) = 3 \times 1.01235 = 3.037 $$
Three-term:
$$ 3 \times (1 + \frac{1}{3} \times 0.03704 - \frac{1}{9} \times 0.03704^2 ) = 3 \times 1.01220 = 3.0366 $$
Actual:
$$ \sqrt[3]{28} = 3.0366 $$
Errors:
Two-term error:
$$ |3.037 - 3.0366| = 0.0004 $$
Three-term error:
$$ |3.0366 - 3.0366| = 0 $$
Answer: Including the third term improves accuracy significantly.
Example 3: Analyze error margin for \(\sqrt{50}\) using two and three terms.
Given: \(a=50\), \(b=49\), \(x=0.02041\).
Two-term:
$$ 7 \times (1 + \frac{1}{2} \times 0.02041) = 7 \times 1.0102 = 7.0714 $$
Three-term:
$$ 7 \times (1 + \frac{1}{2} \times 0.02041 - \frac{1}{8} \times 0.02041^2 ) = 7 \times 1.01015 = 7.07105 $$
Actual:
$$ \sqrt{50} = 7.07107 $$
Errors:
Two-term error:
$$ |7.0714 - 7.07107| = 0.00033 $$
Three-term error:
$$ |7.07105 - 7.07107| = 0.00002 $$
Answer: Third term greatly reduces approximation error.
Example 4: Check convergence for \(\sqrt{18}\) with two terms only.
Given: \(a=18\), \(b=16\), \(x=0.125\).
Two-term:
$$ 4 \times (1 + \frac{1}{2} \times 0.125) = 4 \times 1.0625 = 4.25 $$
Actual:
$$ \sqrt{18} = 4.2426 $$
Error:
$$ |4.25 - 4.2426| = 0.0074 $$
Answer: Two terms give a small error; more terms would improve accuracy.
Example 5: Examine error in cube root approximation of 65 with up to three terms.
Given: \(a=65\), \(b=64\), \(x=0.015625\).
Two-term:
$$ 4 \times (1 + \frac{1}{3} \times 0.015625) = 4 \times 1.005208 = 4.02083 $$
Three-term:
$$ 4 \times (1 + \frac{1}{3} \times 0.015625 - \frac{1}{9} \times 0.015625^2 ) = 4 \times 1.005183 = 4.02073 $$
Actual:
$$ \sqrt[3]{65} = 4.02073 $$
Errors:
Two-term error:
$$ |4.02083 - 4.02073| = 0.0001 $$
Three-term error:
$$ |4.02073 - 4.02073| = 0 $$
Answer: Three terms yield an exact approximation to 5 decimal places.
Quantity surveyors often adjust unit costs or quantities based on changes in dimensions involving roots, such as scaling areas or volumes. Applying binomial expansion to approximate these roots allows quick recalculations without complex tools. This method supports cost estimation and variation analysis in projects where measurements slightly deviate from design values.
Example 1: A floor area increases from 100 m² to 104 m². Estimate the new side length assuming a square floor.
Given: Initial side length \(s_0 = \sqrt{100} = 10\) m, new area \(A = 104\) m².
Calculate \(x = \frac{104 - 100}{100} = 0.04\).
Formula:
$$ s = 10 (1 + 0.04)^{\frac{1}{2}} = 10 [1 + \frac{1}{2}(0.04) - \frac{1}{8}(0.04)^2 ] $$
Substitution:
$$ = 10 (1 + 0.02 - 0.0002) = 10 (1.0198) = 10.198 $$
Answer: New side length \(\approx 10.20\) m
Example 2: A cubic storage tank volume changes from 125 m³ to 130 m³. Estimate new side length.
Given: Initial side length \(s_0 = \sqrt[3]{125} = 5\) m, new volume \(V = 130\) m³.
Calculate \(x = \frac{130 - 125}{125} = 0.04\).
Formula:
$$ s = 5 (1 + 0.04)^{\frac{1}{3}} = 5 [1 + \frac{1}{3}(0.04) - \frac{1}{9}(0.04)^2 ] $$
Substitution:
$$ = 5 (1 + 0.0133 - 0.000178) = 5 (1.0131) = 5.0655 $$
Answer: New side length \(\approx 5.07\) m
Example 3: A square slab's side length reduces from 12 m to 11.5 m. Estimate the percentage change in area.
Given: Initial area \(A_0 = 12^2 = 144\) m², new side length \(s = 11.5\) m.
Calculate \(x = \frac{11.5 - 12}{12} = -0.0417\).
Formula for area ratio:
$$ \frac{A}{A_0} = (1 + x)^2 = 1 + 2x + x^2 $$
Substitution:
$$ = 1 + 2(-0.0417) + (-0.0417)^2 = 1 - 0.0834 + 0.00174 = 0.9183 $$
Percentage change:
$$ (0.9183 - 1) \times 100 = -8.17\% $$
Answer: Area decreases by approximately 8.17%
Example 4: Estimate the new volume of a cubic water tank if side length increases from 3 m by 5%.
Given: Initial volume \(V_0 = 3^3 = 27\) m³, \(x = 0.05\).
Formula:
$$ V = 27 (1 + 0.05)^3 = 27 [1 + 3(0.05) + 3(0.05)^2 + (0.05)^3 ] $$
Substitution:
$$ = 27 (1 + 0.15 + 0.0075 + 0.000125) = 27 (1.157625) = 31.26 $$
Answer: New volume \(\approx 31.26\) m³
Example 5: A concrete slab thickness increases from 0.15 m to 0.16 m. Estimate the percentage increase in volume for a fixed area of 50 m².
Given: Initial volume \(V_0 = 50 \times 0.15 = 7.5\) m³.
Calculate \(x = \frac{0.16 - 0.15}{0.15} = 0.0667\).
Formula for volume ratio:
$$ \frac{V}{V_0} = 1 + x = 1.0667 $$
Percentage increase:
$$ (1.0667 - 1) \times 100 = 6.67\% $$
Answer: Volume increases by approximately 6.67%
Approximate \(\sqrt{18.5}\) using binomial expansion up to the second order term. (5 marks)
Estimate \(\sqrt[3]{65}\) using binomial expansion including terms up to \(x^2\). (6 marks)
Calculate the approximate side length of a square floor with area 110 m² using binomial expansion. (5 marks)
Find the cube root of 1332 m³ by binomial expansion up to the \(x^3\) term. (6 marks)
Determine the error difference between two-term and three-term binomial approximations for \(\sqrt{12}\). (5 marks)
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Create a free accountThis chapter focused on applying the binomial theorem to various engineering mathematics problems. It began with using the binomial theorem to approximate roots of numbers, providing a method to expand expressions involving fractional powers. Next, the chapter explored how to determine binomial series coefficients through Pascal's triangle, demonstrating a systematic approach to find coefficients for any term in the expansion. The final topic addressed the calculation of errors caused by small changes in variables using the binomial theorem, emphasizing its usefulness in estimating variations in engineering measurements. These concepts together form a foundation for expanding and approximating expressions efficiently. The chapter provided a clear progression from understanding the theorem’s structure to practical applications in error estimation. Mastery of these topics prepares students to handle complex expansions and assess the impact of small deviations in engineering calculations.
Calculate the square root of \(81\) using the binomial expansion method by expressing \(\sqrt{81}\) as \(\sqrt{80 + 1}\). (2 marks)
Use Pascal’s triangle to find the coefficients for the expansion of \((1 + x)^4\). Then calculate the value of \((1 + 0.05)^4\) using the binomial expansion. (3 marks)
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