Quadratic equations
Neglecting the negative result which is meaningless, the width of the path, t = 0.651 m or 65 cm, correct to the nearest centimetre.
5. The acid dissociation constant Ka of ethanoic acid is 1.8 × 10−5 mol dm−3 for a particular solution. Using x2 determine x, the the Ostwald dilution law Ka = v(l − x) degree of ionization, given that v = 10 dm3 .
Problem 16. If the total surface area of a solid cone is 486.2 cm2 and its slant height is 15.3 cm, determine its base diameter.
6. A rectangular building is 15 m long by 11 m wide. A concrete path of constant width is laid all the way around the building. If the area of the path is 60.0 m2 , calculate its width correct to the nearest millimetre.
From Chapter 24, page 180, the total surface area A of a solid cone is given by: A = πrl + πr 2 where l is the slant height and r the base radius.
7. The total surface area of a closed cylindrical container is 20.0 m3 . Calculate the radius of the cylinder if its height is 2.80 m2 .
A = 482.2 and l = 15.3, then 482.2 = πr(15.3) + πr 2
If i.e.
π r 2 + 15.3π r − 482.2 = 0
8. The bending moment M at a point in a beam is given 3x(20 − x) where x metres is the distance from by M = 2 the point of support. Determine the value of x when the bending moment is 50 Nm.
482.2 r 2 + 15.3r − =0 π
or
Using the quadratic formula, r= =
−482.2 (15.3)2 − 4 π
−15.3 ±
9. A tennis court measures 24 m by 11 m. In the layout of a number of courts an area of ground must be allowed for at the ends and at the sides of each court. If a border of constant width is allowed around each court and the total area of the court and its border is 950 m2 , find the width of the borders.
2 −15.3 ±
√
848.0461
2
=
−15.3 ± 29.12123 2
10. Two resistors, when connected in series, have a total resistance of 40 ohms. When connected in parallel their total resistance is 8.4 ohms. If one of the resistors has a resistance Rx ohms:
Hence radius r = 6.9106 cm (or −22.21 cm, which is meaningless, and is thus ignored). Thus the diameter of the base
75
= 2r = 2(6.9106)
(a) show that R2x − 40Rx + 336 = 0 and
= 13.82 cm
Now try the following exercise Exercise 40 Further practical problems involving quadratic equations (Answers on page 274) 1. The angle of a rotating shaft turns through in t seconds is given by θ = ωt + 12 αt 2 . Determine the time taken to complete 4 radians if ω is 3.0 rad/s and α is 0.60 rad/s2 .
(b) calculate the resistance of each
10.6 The solution of linear and quadratic equations simultaneously Sometimes a linear equation and a quadratic equation need to be solved simultaneously. An algebraic method of solution is shown in Problem 17; a graphical solution is shown in Chapter 13, page 99.
2. The power P developed in an electrical circuit is given by P = 10I − 8I 2 , where I is the current in amperes. Determine the current necessary to produce a power of 2.5 watts in the circuit.
Problem 17. Determine the values of x and y which simultaneously satisfy the equations: y = 5x − 4 − 2x2 and y = 6x − 7
3. The area of a triangle is 47.6 cm2 and its perpendicular height is 4.3 cm more than its base length. Determine the length of the base correct to 3 significant figures.
For a simultaneous solution the values of y must be equal, hence the RHS of each equation is equated. Thus
4. The sag l metres in a cable stretched between two 12 + x. supports, distance x m apart is given by: l = x Determine the distance between supports when the sag is 20 m.
5x − 4 − 2x2 = 6x − 7 Rearranging gives: i.e. or
5x − 4 − 2x2 − 6x + 7 = 0 −x + 3 − 2x2 = 0 2x2 + x − 3 = 0