Equations in Science
Rule 4: A term can be moved from one side to the other, provided that its sign is reversed Rule 5: A multiplier/divisor factor of one side can become a divisor/multiplier factor of the
3.5 Quadratic and Simultaneous Equations
3.5.1
Introduction
Quadratic and simultaneous equations are two further groups of equations in which we are trying to find the value of unknown variables.
3.5.2
Quadratic equations
A common form of equation where a power is involved is a ‘quadratic equation’. The term
2p2− p − 6 = 0 is a quadratic equation in p.
In general, there will be two possible values ofp that will make a quadratic equation true,
each of which will be a possible solution.
Example 3.41
Check, by substituting values, that both p= 2 and p = −1.5 will make the equation
2p2− p − 6 = 0 true, i.e. 2p2− p − 6 will equal zero.
Putting p= 2: 2× 22− 2 − 6 = 8 − 2 − 6 ⇒ 0 True
Putting p= −1.5: 2× (−1.5)2− (−1.5) − 6 = 2 × 2.25 + 1.5 − 6
⇒ 4.5 + 1.5 − 6 ⇒ 0 True
The general method for finding the solutions starts by rearranging the quadratic equation into a standard form:
ax2+ bx + c = 0 [3.4]
wherea, b and c are constants in the equation, and x is the unknown variable.
With the equation in the above form, the solutions can be found by substituting the values ofa, b and c into the formula:
x =−b ±
√
b2− 4ac
2a [3.5]
The± sign in the numerator of the equation tells us to take two solutions, one with the ‘+’ sign and one with the ‘−’ sign.
It is a straightforward process to use the above formula to solve any quadratic equation: 1. Rearrange the equation into the formax2+ bx + c = 0.
2. Work out the values in the equation equivalent toa, b and c.
Example 3.42
Solve the equation: 2.4x2= 3.2x + 6.6
Rearrange as in [3.4]: 2.4x2− 3.2x − 6.6 = 0
Compare directly with [3.4]: ax2+ bx + c = 0
giving the equivalences: a = 2.4, b = −3.2, c = −6.6
Substitute these values in [3.5]: x =−(−3.2) ±
(−3.2)2− 4 × 2.4 × (−6.6) 2× 2.4 x = +3.2 ± √ 10.24+ 63.36 4.8 ⇒ +3.2 ±√73.6 4.8 ⇒ +3.2 ± 8.58 4.8
giving two solutions forx: x = 2.45 or x = −1.12
Q3.32
Solve the following equations:
(i) 2x2− 3x + 1 = 0 (iii) 4x = 5 − 2x2
(ii) 3.2x2− 2.5x − 0.8 = 0 (iv) (x− 2)2= 2x
When the quadratic equation represents a scientific relationship, then the correct solution in the scientific context may be just one, or both, of the possible mathematical solutions. It is necessary to use other information about the scientific problem to decide which solution (or both) is correct.
Example 3.43
A cricket ball is hit directly upwards with an initial velocity of 16.0 m s−1. The equation for the time,t (seconds), at which it passes a height of 5.0 m is given by:
5.0= 16.0t − 4.9t2
Solving the quadratic equation gives two values,t = 0.35 and 2.92 s.
In the scientific context , the two values are the times (in seconds) at which the ball passes the height 5.0 m, first going up and then coming down!
Example 3.44
The refractive index, n, of a piece of glass is given by solving the equation n(n− 1) = 0.75. Find the value of n.
Expanding the equation givesn2− n = 0.75 or n2− n − 0.75 = 0. This quadratic equation can be solved giving n= 1.5 or n = −0.5.
However, the refractive index of a material is always positive, so in this casen= 1.5 is
the only solution.
In some equations, the value ofb2− 4ac, inside the square root, may be a negative value. It is not normally possible to take the square root of a negative number. Hence:
Ifb2< 4ac the equation has no real solutions [3.6]
Using complex numbers it is possible to derive a ‘square root’ from a negative number, but these techniques are beyond the mathematical scope of this book.
When the value ofb2− 4ac is zero, the square root term also becomes zero:
Ifb2= 4ac, both solutions to the equation have the same value [3.7]
Q3.33
Solve the following equations:
(i) x2− 4x + 4 = 0 (iii) 2x2− 3x + 4 = 0
(ii) x2+ 0.5x − 1.5 = 0 (iv) 4x2+ 12x + 9 = 0
3.5.3
Simultaneous equations
The simplest example of a simultaneous equations problem has: • two equations, which contain
• the same two variables, e.g. x and y, and
• with the requirement that the two equations must both be true (simultaneously) for the same values forx and y.
For example, we may have:
3x+ 2y = 7, and
4x+ y = 6
Both of the above equations are true (simultaneously) ifx= 1 and y = 2. Check by substituting
the values into the equations and finding that they both ‘balance’:
3× 1 + 2 × 2 = 7 True
4× 1 + 2 = 6 True
The solution for this pair of simultaneous equations isx= 1 and y = 2.
If the equations have three variables, e.g. x, y and z, then we will need three equations
to find a solution. In general, if we haven different variables then we will need n different
equations to find solutions for alln variables.
There are several different ways of finding the solution such that all the equations become true simultaneously. Analytical methods will give an exact solution, but may be mathematically difficult to perform. A graphical method can be used to find an (approximate) solution for more complicated equations.
3.5.4
Analytical solution for simultaneous equations
The most reliable analytical method for solving simultaneous equations aims to rearrange the equations so that one of the variables can be ‘eliminated’ from the equation.
The process follows four basic steps, which are illustrated by solving the simultaneous equations in Example 3.45.
Example 3.45
Find values ofp and q that make both of the following equations true:
4p− 2q = 9 [A]
3p+ 7q = 2 [B]
equations. For example, taking the variablep, Equation [A] becomes:
4p= 9 + 2q [C]
and Equation B becomes:
3p= 2 − 7q [D]
Step 2: The aim of this step is to arrive at an equation with only one variable involved. In this example we eliminate p, to get an equation with only q.
To do this, we multiply both sides of [C] by the multiplier ofp(= 3) in [D]:
3× 4p = 3 × 9 + 3 × 2q
12p= 27 + 6q [E]
Similarly we multiply both sides of [D] by the multiplier ofp (= 4) in [C]:
4× 3p = 4 × 2 − 4 × 7q
12p= 8 − 28q [F]
From [E], we see that 27+ 6q equals 12p, and from [F] we see that 8 − 28q also equals 12p. So we can write this equality as a new equation:
27+ 6q = 8 − 28q [G]
Step 3: We can now solve this equation to findq, by rearranging it to make q the subject of
the equation:
6q+ 28q = 8 − 27
34q = −19
Note: We have ‘rounded off’ the value ofq to 5 significant figures. It is important to be careful
about rounding off too much in the middle of a problem in case a subsequent step calculates a small difference between two large numbers.
Step 4: We can now work backwards to findp, by substituting the value of q back into either
[C] or [D]. Using [C]:
4p= 9 + 2 × (−0.55882) ⇒ 7.8824
p= 7.8824/4 ⇒ 1.9706
Using the above four steps, the solutions (to 5 sf) to the equations are:
p= 1.9706 and q = −0.55882
Checking the results by substituting back into the original equations, [A] and [B], to make sure that they are both true:
4× 1.9706 − 2 × (−0.55882) = 9.0000 3× 1.9706 + 7 × (−0.55882) = 2.0001
Allowing for small rounding errors, the calculated values on the RHS agree with the values in the original equations. Hence we have confirmed that the calculated values forp and q are
the solutions to the original equations.
Q3.34
Two walkers, Alex and Ben, start walking towards each other along a path. Alex starts 0.5 km from a village on the path and Ben starts 3.2 km from the village. Alex walks at 1.2 m s−1and Ben walks at 0.9 m s−1.
We can describe the position of each man by giving the distance,d (in metres) of
each man from the village as a function of time,t:
Alex:d = 500 + 1.2 × t
Ben: d= 3200 − 0.9 × t
Solve the simultaneous equations using an analytical method, and calculate (i) The time taken before they meet.
Q3.35
We can sometimes calculate the individual concentrations, C1 and C2, of two compounds in a chemical mixture by measuring the spectrophotometric absorbance,
Aλ, of the mixture at different wavelengths,λ.
The results of a particular experiment produce the following simultaneous equations:
0.773= 2.25 × C1+ 2.0 × C2 0.953= 0.25 × C1+ 6.0 × C2
Use the analytical method to solve the equations and derive the solutions forC1 andC2.
It is possible to use a software procedure in Excel called ‘Solver’ to solve simultaneous equations – see Appendix I and the Website.
3.5.5
Graphical method for simultaneous equations
In the graphical method, the equations are plotted as lines on the same graph. The solution is given by the co-ordinates of the point where the lines cross.
Example 3.46
Solve the simultaneous equations:
y = x2
y = 1.5 − 0.5x The analysis is performed in the following text .
To solve the simultaneous equations from Example 3.46 we plot these equations as lines on a graph by choosing specific values ofx and calculating the equivalent values of y for the two
equations, as in the table below:
x −2 −1 0 1 2
y = x2 4 1 0 1 4
y = 1.5 − 0.5x 2.5 2 1.5 1 0.5
0 1 2 3 4 −2 −1 0 1 2 x y y = x2 y = 1.5 − 0.5 x
Figure 3.1 Simultaneous solutions.
The solutions to the simultaneous equations are given by the two points where the lines cross. The intersection points are identified by two circles that have approximate values (x =
−1.45, y = 2.2) and (x = 1, y = 1). These give two approximate solutions to the equations: Solution 1: x = −1.45 and y = 2.2
Solution 2: x = 1.0 and y = 1.0
There will always be uncertainties in reading the co-ordinates of the crossing point from the graph. However, it is possible, once an approximate value is obtained, to redraw the graph to a much larger scale just around the crossing point.
The particular problem in Example 3.46 can also be solved analytically by first eliminating
y from the two equations, giving
x2= 1.5 − 0.5x
This is now a quadratic equation that can be solved using [3.5] to give two solutions forx – see
Q3.36.
Q3.36
Solve the simultaneous equations (given in Example 3.46)
y = x2
y = 1.5 − 0.5x
using an analytical method.
Q3.37