Quadratic Equations in Engineering

CHAPTER 2

In this chapter, the applications of quadratic equations in engineering are introduced. It is assumed that students are familiar with this topic from their high school algebra course. A quadratic equation is a second-order polynomial equation in one variable that occurs in many areas of engineering. For example, the height of a ball thrown in the air can be represented by a quadratic equation. In this chapter, the solution of quadratic equations will be obtained by three methods: factoring, the quadratic formula, and completing the square.

2.1 A PROJECTILE IN A VERTICAL PLANE

Suppose a ball thrown upward from the ground with an initial velocity of 96 ft/s reaches a height h(t) after time t s as shown in Fig. 2.1. The height is expressed by the quadratic equation h(t) = 96 t − 16 t² ft. Find the time t in seconds when h(t) = 80 ft.

Figure 2.1 A ball thrown upward to a height of h(t).

Solution:

h(t) = 96 t − 16 t² = 80

or

Equation (2.1) is a quadratic equation of the form ax² + bx + c = 0 and will be solved using three different methods.

Method 1: Factoring Dividing equation (2.1) by 16 yields

Equation (2.2) can be factored as

(t − 1)(t − 5) = 0.

Therefore, t − 1 = 0 or t = 1 s and t − 5 = 0 or t = 5 s. Hence, the ball reaches the height of 80 ft at 1 s and 5 s.

Method 2: Quadratic Formula If ax² + bx + c = 0, then the quadratic formula to solve for x is given by

Using the quadratic formula in equation (2.3), the quadratic equation (2.2) can be solved as

Therefore, and . Hence, the ball reaches the height of 80 ft at 1 s and 5 s.

Method 3: Completing the Square First, rewrite the quadratic equation (2.2) as

Adding the square of (one-half the coefficient of the first-order term) to both sides of equation (2.4) gives

or

Equation (2.5) can now be written as

or

t − 3 = ±2.

Therefore, t = 3 ± 2 or t = 1, 5 s. To check if the answer is correct, substitute t = 1 and t = 5 into equation (2.1). Substituting t = 1 s gives

16 × 1 − 96 × 1 + 80 = 0,

which gives 0 = 0. Therefore, t = 1 s is the correct time when the ball reaches a height of 80 ft. Now, substitute t = 5 s,

16 × 5² − 96 × 5 + 80 = 0,

which again gives 0 = 0. Therefore, t = 5 s is also the correct time when the ball reaches a height of 80 ft.

It can be seen from Fig. 2.2 that the height of the ball is 80 ft at both 1 s and 5 s. The ball is at 80 ft and going up at 1 s, and it is at 80 ft and going down at 5 s. Hence, the maximum height of ball must be halfway between 1 and 5 s, which is 1 + ((5 − 1)/2) = 3 s. Therefore, the maximum height can be found by substituting t = 3 s in h(t), which is h(3) = 96(3) − 16(3)² = 144 ft. These three points (height at t = 1, 3, and 5 s) can be used to plot the trajectory of the ball. However, to plot the trajectory accurately, additional data points can be added. The height of the ball at t = 0 is zero since the ball is thrown upward from the ground. To check this, substitute t = 0 in h(t). This gives h(0) = 96(0) − 16(0)² = 0 ft. The time when the ball hits the ground again can be calculated by equating h(t) = 0. Therefore,

Therefore, t = 0 and 6 − t = 0 or t = 6 s. Since the ball is thrown in the air from the ground (h(t) = 0) at t = 0, it will hit the ground again at t = 6 s. Using these data points, the trajectory of the ball thrown upward with an initial velocity of 96 ft/s is shown in Fig. 2.2.

Figure 2.2 The height of the ball thrown upward with an initial velocity of 96 ft/s.

Suppose now you wish to find the time t in seconds when the height of the ball reaches 144 ft. Setting h(t) = 144 gives

h(t) = 96 t − 16 t² = 144.

Therefore,

16 t² − 96 t + 144 = 0

or

The quadratic equation given in equation (2.6) can also be solved using the three methods as

Now suppose you wish to find the time t when the height of the ball reaches h(t) = 160 ft. Setting h(t) = 160 gives

h(t) = 96t − 16t² = 160.

Therefore,

16t² − 96t + 160 = 0

or

The quadratic equation given in equation (2.7) can be solved using the three methods as

In the above solution i = j = is the imaginary number, therefore the roots of the quadratic equation are complex. Hence, the ball never reaches the height of 160 ft. The maximum height achieved is 144 ft at 3 s.

2.2 CURRENT IN A LAMP

A 100 W lamp and a 20 Ω resistor are connected in series to a 120 V power supply as shown in Fig. 2.3. The current I in amperes satisfies a quadratic equation as follows. Using KVL,

120 = V_L + V_R.

Figure 2.3 A lamp and a resistor connected to a 120 V supply.

From Ohm's law, V_R = 20 I. Also, since the power is the product of voltage and current, P_L = V_L I = 100 W, which gives . Therefore,

Multiplying both sides of equation (2.8) by I yields

Dividing both sides of equation (2.9) by 20 and rearranging gives

The quadratic equation given in equation (2.10) can be solved using the three methods as

Note that the two solutions correspond to two lamp choices.

Case I: For I = 1 A,

Case II: For I = 5 A,

Case I corresponds to a lamp rated at 100 V, and Case II corresponds to a lamp rated at 20 V.

2.3 EQUIVALENT RESISTANCE

Suppose two resistors are connected in parallel, as shown in Fig. 2.4. If the equivalent resistance and R₁ = 4R₂ + 100 Ω, find R₁ and R₂.

Figure 2.4 Equivalent resistance of two resistors connected in parallel.

The equivalent resistance of two resistors connected in parallel as shown in Fig. 2.4 is given by

Substituting R₁ = 4R₂ + 100 Ω in equation (2.11) gives

Multiplying both sides of equation (2.12) by 5R₂ + 100 yields

Simplifying equation (2.13) gives

Dividing both sides of equation by (2.14) by 4 gives

Equation (2.15) is a quadratic equation in R₂ and cannot be factored with whole numbers. Therefore, R₂ is solved using the quadratic formula as

Therefore,

Since R₂ cannot be negative,

Substituting the value of R₂ in R₁ = 4R₂ + 100 Ω yields

R₁ = 4(120.7) + 100 = 582.8 Ω.

Therefore, R₁ = 582.8 Ω and R₂ = 120.7 Ω.

2.4 FURTHER EXAMPLES OF QUADRATIC EQUATIONS IN ENGINEERING

Solution

(a) Substituting h(t) = 245 in equation (2.16), the quadratic equation is given by

Dividing both sides of equation (2.17) by −4.9 gives

The quadratic equation given in equation (2.18) can be solved using the three methods used in Section 2.1 as

(b) Since the rocket hit the ground at h(t) = 0,

Therefore, t = 0 s and t = 20 s. Since the rocket is fired from the ground at t = 0 s, the rocket hits the ground again at t = 20 s.

(c) The maximum height should occur halfway between 2.93 and 17.07 s. Therefore,

Substituting t = 10 s into equation (2.16) yields

h_max = 98(10) − 4.9(10)² = 490 m.

The plot of the rocket trajectory is shown in Fig. 2.6. It can be seen from this figure that the rocket is fired from the ground at a height of zero at 0 s, crosses a height of 245 m at 2.93 s, and continues moving up and reaches the maximum height of 490 m at 10 s. At 10 seconds, it starts its downward descent and after crossing the height of 245 m again at 17.07 s, it reaches the ground again at 20 s

Figure 2.6 The height of the rocket fired vertically in the air with an initial velocity of 98 m/s.

Solution

(a) Substituting R₂ = 2R₁ + 4 and R = 8.0 in equation (2.19) gives

Multiplying both sides of equation (2.20) by (3R₁ + 4) yields

or

Rearranging terms in equation (2.21) gives

(b) The quadratic equation given in equation (2.22) can be now be solved to find the values of R₁.

(i) Method 1: Completing the square:

Dividing both sides of equation (2.22) by 2 gives

Taking 16 on the other side of equation (2.23) and adding to both sides yields

Now, writing both sides of equation (2.24) as squares yields

Therefore,

R₁ − 5 = ±6.4,

which gives the values of R₁ as 5 + 6.4 = 11.4 Ω and 5 − 6.4 = −1.4 Ω. Since the value of R₁ cannot be negative, R₁ = 11.4 Ω and R₂ = 2R₁ + 4 = 2(11.4) + 4 = 26.8 Ω.

(ii) Method 2: Solving equation (2.22) using the quadratic formula:

Since R₁ cannot be negative, R₁ = 11.4 Ω. Substituting R₁ = 11.4 Ω in R₂ = 2 R₁ + 4 gives

R₂ = 2(11.4) + 4 = 26.8 Ω.

Solution

(a) Substituting k₂ = 2k₁ + 4 and k = 3.6 in equation (2.25) yields

Multiplying both sides of equation (2.27) by (3k₁ + 4) gives

Rearranging terms in equation (2.28) gives

(b) The quadratic equation (2.29) can be solved using the quadratic formula as

Since k₁ cannot be negative, k₁ = 2.0 lb/in. Now, substituting k₁ = 2.0 in k₂ = 2k₁ + 4 yields

k₂ = 2(2) + 4 = 8.0.

Therefore,

k₂ = 8.0 lb/in.

Solution

(a) The total reactance of the series combination of L and C shown in Fig. 2.8 is given by

Substituting L = 1.0 H, C = 0.25 F, and X = 3.0 Ω in equation (2.30) yields

Multiplying both sides of equation (2.31) by ω gives

Rearranging terms in equation (2.32) yields

(b) The quadratic equation (2.33) can be solved by three different methods: factoring, completing the squares, and the quadratic formula.

(i) Method 1: Factoring:

The quadratic equation (2.33) can be factored as

(ω − 4)(ω + 1) = 0,

which gives ω − 4 = 0 or ω + 1 = 0. Therefore, ω = 4 rad/s or ω = −1 rad/s. Since ω cannot be negative, ω = 4 rad/s.

(ii) Method 2: Completing the squares:

The quadratic equation (2.33) can be written as

Adding to both sides of equation (2.34) gives

Therefore,

Writing both sides of equation (2.35) as a square gives

Taking the square root of both sides of equation (2.36) yields

Therefore,

which gives rad/s or rad/s. Since ω cannot be negative, ω = 4 rad/s.

(iii) Method 3: Quadratic formula:

Solving the quadratic equation (2.33) using the quadratic formula gives

Equation (2.37) can be written as

which gives ω = 4, −1. Since ω cannot be negative, ω = 4 rad/s.

Solution

(a) Substituting P = 96 W, V_L = 32 V, and R = 8 Ω into the power delivered P = I² R + I V_L yields

Dividing both sides of equation (2.38) by 8 gives

Rearranging terms in equation (2.39) yields

(b) The quadratic equation given in equation (2.40) can be solved by three different methods: factoring, completing the squares, and the quadratic formula.

(i) Method 1: Factoring:

The quadratic equation (2.40) can be factored as

(I + 6)(I − 2) = 0,

which gives I + 6 = 0 or I − 2 = 0. Therefore, I = −6 A or I = 2 A.

(ii) Method 2: Completing the square:

The quadratic equation (2.40) can be written as

Adding to both sides of equation (2.41),

Writing both sides of equation (2.42) as a square yields

Taking the square root of both sides of equation (2.43) gives

I + 2 = ±4.

Therefore,

I = −2 ±4,

which gives I = −2 −4 = −6 A or I = −2 + 4 = 2 A

(iii) Method 3: Quadratic formula:

Solving the quadratic equation (2.40) using the quadratic formula gives

Equation (2.44) can be written as

which gives I = −2 − 4 = −6 A or I = −2 + 4 = 2 A.

Case I: For I = −6 A, the power absorbed by the lamp is −6 × 32 = −192 W. Since the power absorbed by the lamp cannot be negative, I = −6 A is not one of the solutions of the quadratic equation given by (2.40).

Case II: For I = 2 A, the power absorbed by the lamp is 2 × 32 = 64 W and the power dissipated by the resistor is 96 − 64 = 32 W. The voltage across the resistor V_R = 2 × 8 = 16 V and using KVL, V_s = 16 + 32 = 48 V. Therefore, for the applied power of 96 W (source voltage = 48 V), I = 2 A is the solution of the quadratic equation given by (2.40).

Solution

(a) The time when the diver hits the water is found by setting h(t) = 0 in equation (2.45) as

Dividing both sides of equation (2.46) by −4.905 gives

The quadratic equation given in equation (2.47) can be solved using the quadratic formula and completing the square as outlined below.

(i) Method 1: Quadratic formula:

Solving the quadratic equation (2.47) using the quadratic formula gives

Equation (2.48) can be written as

which gives t = 0.0612 − 0.5563 = −0.495 s or t = 0.0612 + 0.5564 = 0.617 s. Since the time cannot be negative, it takes 0.617 s for the diver to hit the water.

(ii) Method 2: Completing the square:

The quadratic equation (2.47) can be written as

Adding to both sides gives

Writing both sides of equation (2.50) as a perfect square yields

Taking the square root of both sides gives

t − 0.0612 = ±0.5563.

Therefore,

t = 0.0612 ±0.5563,

which gives t = 0.0612 − 0.5563 = −0.495 s or t = 0.0612 + 0.5563 = 0.617 s. Since the time cannot be negative, it takes 0.617 s for the diver to hit the water.

(b) The maximum height of the diver is found by substituting t = 0.0612 in equation (2.45) as

h_max = h(0.0612) = −4.905(0.0612)² + 0.6(0.0612) + 1.5 = 1.518 m.

(c) Using the results of parts (a) and (b), the diver's height after jumping from the diving board is plotted as shown in Fig. 2.10.

Figure 2.10 Height of the diver after jumping from the diving board.

Solution

(a) Since the height y of the tunnel is 2 m, equation (2.52) can be written as

2 = −0.004x² + 0.3x

which gives

0.004x² − 0.3x + 2 = 0

or

(b) The quadratic equation given in equation (2.53) can be solved using the quadratic formula and completing the square as outlined below.

(i) Method 1: Quadratic formula:

Solving the quadratic equation (2.53) using the quadratic formula gives

Equation (2.54) can be written as

which gives x = 37.5 − 30.1 = 7.4 m or x = 37.5 + 30.1 = 67.6 m. Therefore, the tunnel entry position is 7.4 m and the tunnel exit position is 67.6 m.

(ii) Method 2: Completing the square:

The quadratic equation (2.53) can be written as

Adding to both sides gives

Writing both sides of equation (2.56) as a perfect square yields

Taking the square root of both sides gives

x − 37.5 = ±30.1.

Therefore,

x = 37.5 ± 30.1,

which gives x = 37.5 − 30.1 = 7.4 m or x = 37.5 + 30.1 = 67.6 m. To check if the answer is correct, substitute x = 7.4 and x = 67.6 into equation (2.53). Substituting x = 7.4 m gives

Now, substituting x = 67.6 m into equation (2.53) gives

Therefore, x_A = 7.4 m and x_B = 67.6 m are the correct positions of the tunnel entry and exit, respectively.

(c) The tunnel length can be found by subtracting the position x_A from x_B as

Tunnel length = x_B − x_A = 67.6 − 7.4 = 60.2 m

PROBLEMS

1. 2-1. An analysis of a circuit shown in Fig. P2.1 yields the quadratic equation for the current I as 3I² + 6 I = 45, where I is in amps.

   (a) Rewrite the above equation in the form I² + aI + b = 0, where a and b are constants.

   (b) Solve the equation in part (a) by each of the following methods: factoring, completing the square, and the quadratic formula.

   

   Figure P2.1 Resistive circuit for problem P2-1.

2. 2-2. Repeat problem P2-1 for the circuit shown in Fig. P2.2, which yields the quadratic equation for the current I as 3I² − 6 I = 45, where I is in amps.

   

   Figure P2.2 Resistive circuit for problem P2-2.

3. 2-3. Repeat problem P2-1 for the circuit shown in Fig. P2.3, which yields the quadratic equation for the current I as 210 = 10 I² + 40 I, where I is in amps.

   

   Figure P2.3 Resistive circuit for problem P2-3.

4. 2-4. The current flowing through the inductor shown in Fig. P2.4 is given by the quadratic equation i(t) = t² − 8t. Find t when

   (a) i(t) = 9 A (use the quadratic formula), and

   (b) i(t) = 84 A (use completing the square).

   

   Figure P2.4 Current flowing through an inductor.

5. 2-5. The voltage across the capacitor shown in Fig. P2.5 is given by the quadratic equation v(t) = t² − 6t. Find t when

   (a) v(t) = 16 V (use the quadratic formula), and

   (b) v(t) = 27 V (use completing the square).

   

   Figure P2.5 Voltage across a capacitor.

6. 2-6. In the purely resistive circuit shown in Fig. P2.6, the total resistance R of the circuit is given by

   

   If the total resistance of the circuit is R = 100 Ω and R₂ = 2R₁ + 100 Ω, find R₁ and R₂ as follows:

   (a) Substitute the values of R and R₂ into equation (2.58), and simplify the resulting expression to obtain a single quadratic equation for R₁.

   (b) Using the method of your choice, solve the quadratic equation for R₁ and compute the corresponding value of R₂.

   

   Figure P2.6 Series parallel combination of resistors.

7. 2-7. Repeat problem P2-6 if R = 1600 Ω and R₂ = R₁ + 500 Ω.

8. 2-8. The energy dissipated by a resistor shown in Fig. P2.8 varies with time t in s according to the quadratic equation W = 3t² + 6t. Solve for t if

   (a) W = 3 joules

   (b) W = 9 joules

   (c) W = 45 joules

   

   Figure P2.8 Resistive circuit for problem P2-8.

9. 2-9. The equivalent capacitance C of two capacitors connected in series as shown in Fig. P2.9 is given by

   

   If the total capacitance is C = 120 μF and C₂ = C₁ + 100 μF, find C₁ and C₂ as follows:

   (a) Substitute the values of C and C₂ in equation (2.59) and obtain the quadratic equation for C₁.

   (b) Solve the quadratic equation for C₁ obtained in part (a) by each of the following methods: factoring, completing the square, and the quadratic formula. Also, compute the corresponding values of C₂.

   

   Figure P2.9 Series combination of two capacitors.

10. 2-10. Repeat problem P2-9 if C = 75 μF and C₂ = C₁ + 200 μF.

11. 2-11. The equivalent capacitance C of three capacitors connected in series-parallel as shown in Fig. P2.11 is given by

    

    If the total capacitance is C = 125 μF and C₂ = C₁ + 100 μF, find C₁ and C₂ as follows:

    (a) Substitute the values of C and C₂ in equation (2.60) and obtain the quadratic equation for C₁.

    (b) Solve the quadratic equation for C₁ obtained in part (a) by completing the square and the quadratic formula. Also, compute the corresponding values of C₂.

    

    Figure P2.11 Series combination of two capacitors.

12. 2-12. The equivalent inductance L of two inductors connected in parallel as shown in Fig. P2.12 is given by

    

    If the total inductance L = 80 mH and L₁ = L₂ + 300 mH, find L₁ and L₂ as follows:

    (a) Substitute the values of L and L₁ in equation (2.61) and obtain the quadratic equation for L₂.

    (b) Solve the quadratic equation for L₂ obtained in part (a) by completing the square, and the quadratic formula. Also, compute the corresponding values of L₁.

    

    Figure P2.12 Parallel combination of two inductors.

13. 2-13. Repeat problem P2-12 if L = 150 mH and L₁ = L₂ + 400 mH.

14. 2-14. The equivalent inductance L of three inductors connected in series-parallel as shown in Fig. P2.14 is given by

    

    (a) Suppose L₂ = L₁ + 200 mH and that the equivalent inductance is L = 200 mH. Substitute these values in equation (2.62) and obtain the following quadratic equation:

    

    (b) Solve the quadratic equation (2.63) for L₁ by each of the following methods: factoring, completing the square, and the quadratic formula.

    

    Figure P2.14 Series-parallel combination of three inductors.

15. 2-15. A model rocket is launched in the vertical plane at time t = 0 s as shown in Fig. P2.15. The height of the rocket (in feet) satisfies the quadratic equation h(t) = 64 t − 16 t².

    (a) Find the value(s) of the time t when h(t) = 48 ft.

    (b) Find the value(s) of the time t when h(t) = 60 ft.

    (c) Find the time required for the rocket to hit the ground.

    (d) Based on your solution to parts (a) through (c), determine the maximum height of the rocket and sketch the height h(t).

    

    Figure P2.15 A model rocket for problem P2-15.

16. 2-16. The ball shown in Fig. P2.16 is dropped from a height of 1000 meters. The ball falls according to the quadratic equation h(t) = 1000 − 4.905t². Find the time t in seconds for the ball to reach a height h(t) of

    (a) 921.52 m

    (b) 686.08 m

    (c) 509.5 m

    (d) 0 m

    

    Figure P2.16 A ball dropped from a height of 1000 m.

17. 2-17. At time t = 0, a ball is thrown vertically from the top of the building at a speed of 56 ft/s, as shown in Fig. P2.17. The height of the ball at time t is given by

    h(t) = 32 + 56t − 16t² ft.

    

    Figure P2.17 A ball thrown vertically from the top of a building.

    (a) Find the value(s) of the time t when h(t) = 32 ft.

    (b) Find the time required for the ball to hit the ground.

    (c) Use the results to determine the maximum height, and sketch the height h(t) of the ball.

18. 2-18. Two springs connected in series shown in Fig. P2.18 can be represented by a single equivalent spring. The stiffness of the equivalent spring is given by

    

    where k₁ and k₂ are the spring constants of the two springs. If k_eq = 1.2 N/m and k₂ = 2k₁ − 1 N/m, find k₁ and k₂ as follows:

    (a) Substitute the values of k_eq and k₂ in equation (2.64) and obtain the quadratic equation for k₁.

    (b) Solve the quadratic equation for k₁ obtained in part (a) by completing the square, and the quadratic formula. Also, compute the corresponding values of k₂.

    

    Figure P2.18 Series combination of two springs.

19. 2-19. The equivalent stiffness of a series-parallel combination of three springs shown in Fig. P2.19 is given by

    

    If k₁ = k₂ + 2 lb/in and the equivalent stiffness is k = 1.75 lb/in, find k₁ and k₂ as follows:

    (a) Substitute the values of k and k₁ into equation (2.65) and obtain the quadratic equation for k₂.

    (b) Using the method of your choice, solve the quadratic equation obtained in part (a) and determine the values of k₂. Also, find the values of k₁.

    

    Figure P2.19 Series-parallel combination of three springs.

20. 2-20. An assembly of three springs connected in series as shown in Fig. P2.20 has an equivalent stiffness k given by

    

    (a) Suppose k₂ = 6 lb/in., k₃ = k₁ + 8 lb/in., and the equivalent stiffness is k = 2 lb/in. Substitute these values into equation (2.66) to obtain the following quadratic equation:

    

    (b) Solve equation (2.67) for k₁ by each of the following methods: (i) factoring, (ii) quadratic formula, and (iii) completing the square. For each case, determine the value of k₃ corresponding to the only physical solution for k₁.

    

    Figure P2.20 Series combination of three springs.

21. 2-21. Consider a capacitor C and an inductor L connected in parallel, as shown in Fig. P2.21. The total reactance X in ohms is given by , where ω is the angular frequency in rad/s.

    (a) Suppose L = 1.0 mH and C = 1 F. If the total reactance is X = 1.0 Ω, show that the angular frequency ω satisfies the quadratic equation ω² + ω − 1000 = 0.

    (b) Solve the quadratic equation for ω by the methods of completing the square and the quadratic formula.

    

    Figure P2.21 Parallel connection of L and C.

22. 2-22. Assume that the total reactance in problem P2-21 is X = −1 Ω. Show that the angular frequency ω satisfies the quadratic equation ω² − ω − 1000 = 0, and find the value of ω using both the quadratic formula and completing the square.

23. 2-23. If L = 0.5 H, C = 0.005 F, and the total reactance in problem P2-21 is X = Ω, show that the angular frequency ω satisfies the quadratic equation ω² − 30ω − 400 = 0. Also, find the value of ω using both the quadratic formula and completing the square.

24. 2-24. When converting resistances connected in a Δ formation to a Y formation as shown in Fig. P2.24, the resistance R₁ is obtained as

    

    (a) Suppose R₁ = 100, R_a = R_b = R, and R_c = 100 + R, all measured in ohms. Substitute these values into equation (2.68) to obtain the following quadratic equation for R.

    R² − 300 R − 10,000 = 0

    (b) Solve the quadratic equation for R by the methods of completing the square and the quadratic formula.

    

    Figure P2.24 Delta to Y conversion circuit.

25. 2-25. When converting resistances connected in a Δ formation to a Y formation as shown in Fig. P2.24, the resistance R₁ is given by equation (2.68).

    (a) Suppose R₁ = 6, R_a = R, R_b = 3 R, and R_c = 100 + R, all measured in ohms. Substitute these values into equation (2.68) to obtain the following quadratic equation for R.

    R² − 10R − 200 = 0

    (b) Solve the quadratic equation for R by the methods of completing the square and the quadratic formula.

26. 2-26. When converting resistances connected in a Δ formation to a Y formation as shown in Fig. P2.24, the resistance R₂ is obtained as

    

    (a) Suppose R₂ = 12, R_a = R, R_b = 3 R, and R_c = 100 + R, all measured in ohms. Substitute these values into equation (2.69) to obtain the following quadratic equation for R.

    R² + 40R − 1200 = 0

    (b) Solve the quadratic equation for R by the methods of completing the square and the quadratic formula.

27. 2-27. When converting resistances connected in a Δ formation to a Y formation as shown in Fig. P2.24, the resistance R₃ is obtained as

    

    (a) Suppose R₃ = 36, R_a = R, R_b = 3R, and R_c = 100 + R, all measured in ohms. Substitute these values into equation (2.70) to obtain the following quadratic equation for R.

    R² + 40R − 1200 = 0

    (b) Solve the quadratic equation for R by the methods of completing the square and the quadratic formula.

28. 2-28. The characteristic equation of a series RLC circuit shown in Fig. P2.28 is given as

    

    (a) If R = 7 Ω, L = 1 H, and C = 0.1 F, solve the quadratic equation (2.71) for the values of s (called the eigenvalues of the system) using the methods of completing the square and the quadratic formula.

    (b) Repeat part (a) if R = 10 Ω, L = 1 H, and F.

    

    Figure P2.28 Series RLC circuit for problem P2-28.

29. 2-29. The characteristic equation of a parallel RLC circuit shown in Fig. P2.29 is given as

    

    If R = 200 Ω, L = 50 mH, and C = 0.2 μF, solve the quadratic equation (2.72) for the values of s by the methods of completing the square and the quadratic formula.

    

    Figure P2.29 Parallel RLC circuit for problem P2-29.

30. 2-30. The characteristic equation of a mass, spring, and damper system shown in Fig. P2.30 is given by

    

    (a) If m = 1 kg, c = 3 Ns/m, and k = 2 N/m, solve the quadratic equation (2.73) for the values of s using the methods of completing the square and the quadratic formula.

    (b) Repeat part (a) if m = 1 kg, c = 2 Ns/m, and k = 1 N/m.

    

    Figure P2.30 Mass, spring, and damper system for problem P2-30.

31. 2-31. The perimeter of a rectangle shown in Fig. P2.31 is given by

    

    

    Figure P2.31 A rectangle of length L and width W.

    If the perimeter P = 30 m and the area A = W × L = 36 m², find the length L and width W as follows:

    (a) Substitute the values of P and A in equation (2.74) and obtain the quadratic equation for L.

    (b) Solve the quadratic equation for L obtained in part (a) by completing the square, and the quadratic formula. Also, compute the corresponding values of W.

32. 2-32. A diver jumps off a diving board 2.0 m above the water with an initial vertical velocity of 0.981 m/s as shown in Fig. P2.32. The height h(t) above the water is given by

    h(t) = −4.905 t² + 0.981t + 2.0 m.

    (a) Find the time in seconds when the diver hits the water. Use both the quadratic formula and completing the square.

    (b) Find the maximum height of the diver if it is known to occur at t = 0.1 s.

    (c) Use the results of parts (a) and (b) to sketch the height h(t) of the diver.

    

    Figure P2.32 Diver jumping off a diving board.

33. 2-33. A level pipeline is required to pass through a hill having a parabolic profile

    

    The origin of the x- and y-coordinates is fixed at elevation zero near the base of the hill, as shown in Fig. P2.33.

    

    Figure P2.33 Pipeline path through a parabolic hill.

    (a) Write the quadratic equation for a pipeline elevation of y = 3 m.

    (b) Solve the quadratic equation found in part (a) to determine the positions of the tunnel entry x_A and exit x_B using both the quadratic formula and completing the square.

    (c) Find the length of the tunnel.

34. 2-34. A research group is using a drop test to measure the force of attenuation of a helmet liner they designed to reduce the occurance of brain injuries for soldiers and athletes. The helmet attached to a weight is propelled downward with an initial velocity v_i of 3 m/s from an initial height of 30 m. The behavior of the falling helmet is characterized by a quadratic equation h(t) = 30 − 3 t − 4.9 t².

    (a) Write the quadratic equation for time t when the helmet and weight hit the ground, i.e., h(t) = 0 m.

    (b) Solve the quadratic equation for t obtained in part (a) by completing the square, and the quadratic formula.

35. 2-35. The modulus of elasticity (E) is a measure of a material's resistance to deformation; the larger the modulus, the stiffer the material. During fabrication of a ceramic material from a powder form, pores were generated that affect the strength of the material. The magnitude of elasticity is related to volume fraction porosity P by

    

    (a) If a porous sample of silicon nitride has a modulus of elasticity of E = 209 GPa, obtain the quadratic equation for volume fraction porosity P.

    (b) Solve the quadratic equation found in part (a) using both the methods of completing the square and the quadratic formula.

    (c) Repeat parts (a) and (b) to find the volume fraction porosity of a silicon nitride sample with a modulus of elasticity of 25 GPa.

36. 2-36. Consider the following reaction having an equilibrium constant of 4.66 × 10⁻³ at a certain temperature:

    A(g) + B(g) ⇌ C(g)

    If 0.300 mol of A and 0.100 mol of B are mixed in a container and allowed to reach equilibrium, the concentrations of A = 0.300 − x and B = 0.100 − x reaction that form the concentration of C = 2x are related to the equilibrium constant by the expression

    .

    (a) Write the quadratic equation for x.

    (b) Solve the quadratic equation found in part (a) by completing the square and the quadratic formula.

    (c) Find the equilibrium concentration A, B, and C.

37. 2-37. Consider the following reaction having an equilibrium constant of 1.00 at 1105°K temperature:

    CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g)

    Suppose the feed to a reactor contains 1.000 mol of CO and 2.000 mol of H₂O(g), and the reaction mixture comes to equilibrium at 1105 K. The concentrations of CO = 1.000 − x and H₂O = 2.000 − x reaction that form the concentration of CO₂ = x and H₂ = x are related to the equilibrium constant as

    

    (a) Write the quadratic equation for x.

    (b) Solve the quadratic equation found in part (a) by completing the square and the quadratic formula.

    (c) Find the equilibrium concentration of CO, H₂O, CO₂, and H₂.

38. 2-38. An engineering co-op wants to hire an asphalt contractor to widen the truck entrance to the corporate headquarters as shown in Fig. P2.38.

    

    Figure P2.38 New asphalt dimensions of the corporate headquarters driveway.

    The height h is required to be 10 ft. more than the width b, and the total area of the new asphalt is given by

    

    (a) If A = 200 sq. ft., obtain the quadratic equation for b.

    (b) Solve the quadratic equation found in part (a) by completing the square and the quadratic formula.

    (c) Find the dimensions of the width and height.

39. 2-39. Repeat problem P2-38 if the total area of the new asphalt needs to be 300 sq. ft.

40. 2-40. A city wants to hire a contractor to build a walkway around the swimming pool in one of its parks. The dimensions of the walkway along with the dimensions of the pool are shown in Fig. P2.40. The area of the walkway is given by

    A = (50 + 5 x)(30 + 2 x) − 1500

    (a) If A = 4500 sq. ft., obtain the quadratic equation for x.

    (b) Solve the quadratic equation found in part (a) by completing the square and the quadratic formula.

    

    Figure P2.40 Walkway around the swimming pool.