Section B Algebra
5.1 Introduction A linear equation is one where the highest power of a variable is 1 (see chapter 4). Such equations have the general form
ax + b = 0
where x is the variable and a and b are constants. There is just one solution, x = -b/a.
The term quadratic equation is used for an equation where the highest power of a variable is 2. Such equations have the general form
ax2 + bx + c = 0
where x is the variable and a, b and c are constants. Such equations have two solutions and the obtaining of such solutions is discussed in this chapter.
Quadratic equations occur often in engineering. An example of such an equation in engineering occurs with uniformly accelerated motion in a straight line, namely
s = ut + ½at2
where s is the distance travelled in time t, u the initial velocity and a the acceleration. This equation can be rewritten in the general form indicated above, of ax2 + bx + c = 0, as:
½at2 + ut - s = 0
The linear equation and the quadratic equation are just two examples of what are termed polynomials. A polynomial is the term used for any equation involving powers of the variable which are positive integers. Such powers can be 1, 2, 3, 4, 5, etc. For example, x4 + 4x3 + 2x2 + 5x + 2 = 0 is a polynomial with the highest power being 4. In this chapter the discussion is restricted to just the quadratic equation. The constants a, b, c, etc. that are used to multiply the x terms are termed coefficients. Thus we might have with a polynomial having the highest power 3:
ax3 + bx2 + cx1 + dx0 = 0 Example
Which of the following are quadratic equations?
(a) 2x + 6 = 0, (b) x2 + 3x + 2 = 0, (c) 3x = 5, (d) x2 = 4 + 2x,
(e) 2x3 + 5x2 + 3x + 1 = 0, (f) x2 = 20, (g) 1
X + 2 x = 4.
Equations (a) and (c) are linear equations since they can be written in the form ax + b = 0. Thus (c) can be written as 3x - 5 = 0. Equations (b), (d), (f) and (g) are quadratic equations since they can be written in the form ax2 + bx + c = 0, even if b or c are zero (a must not be zero if it is to be a quadratic). Equation (d) can be written as x2 - 2x - 4 = 0 and equation (f) as x2 + 0x - 20 = 0. When both sides of equation (g) are multiplied by x we obtain 1 + 2x2 = 4x which rearranges to give the equation 2x2 - 4x + 1 = 0. Equation (e) has the highest power of the variable as 3 and so is not a linear or quadratic equation. All the equations are polynomials.
Revision
1 Which of the following equations are (i) linear equations, (ii) quadratic equations?
(a) x = 3, (b) 2x + 5 = 0, (C) x2 = 4, (d) x2 + 5x + 2 = 0, (e) x3 + x2 = 4, (f) 4x2 + 3x = 5, (g) 1
x + 1 = 3, (h) 1
x - 3 x = 5.
5.2 Factors To factorise a number means to write it as the product of smaller numbers.
Thus, for example, we can factor 12 to give 12 = 3 x 4. The 3 and 4 are factors of 12. If the 3 and the 4 are multiplied together then 12 is obtained.
To factorise a polynomial means to write it as the product of simpler poly-nomials. Thus for the quadratic expression x2 + 5x + 6 we can write:
x2 + 5x + 6 = (x + 2)(x + 3)
(x + 2) and (x + 3) are factors. If the two factors are multiplied together then the x2 + 5x + 6 is obtained. Note that, in general:
(a + b)(c + d) = a(c + d) + b(c + d) = ac + ad + bc + bd For example:
(x + 2)(x + 3) = x(x + 3) + 2(x + 3) = x2 + 3x + 2x + 6 = x2 + 5x + 6 Note that when we multiply the first terms in each bracket we obtain the first term of the quadratic expression. When the last term in each bracket are multiplied we obtain the last term of the quadratic expression. Thus, in general, we look for the factors of a quadratic expression by finding all the possible factors of the first and last terms and finding which combination will give the correct middle term.
The following are some commonly encountered forms of equations and their factors:
x2 + (a + b)x + ab (x + a)(x + b) [1]
x2 + 2ax + a2 (x + a)(x + a) or (x + a)2 [2]
x2 - 2ax + a2 (x - a)(x - a) or (x - a)2 [3]
x2 - a2 (x + a)(x - a) [4]
x2 + ax x(x + a) [5]
Example
Factorise: (a) x2 + 4x + 3, (b) 2x2 + 13x + 15, (c) x2 + 4x, (d) x2 - 25.
(a) The first term has the only two factors of lx and lx. The last term has the only two factors of 3 and 1. Thus the factors are (x + 3) and (x + 1), the product of these two giving the correct middle term.
(b) The first term has the only two factors of 2x and lx. The last term has the only two factors of 5 and 3. Combinations of these factors will give (2x + 5)(x + 3), (2x + 3)(x + 5). The product of the two bracketed terms are 2x2 + 11x + 15 and 2x2 + 13x + 15. Thus the correct factors are (2x + 3)(x + 5).
(c) We can extract an x from both terms in the expression to give x(x + 4) and so these are the factors.
(d) Since 25 is 52 we have an expression which is the difference of two squares and so of the form given in equation [4] above. Thus the factors are (x + 5)(x - 5).
Revision
1 Factorise:
(a) x2 - 3x + 2, (b) x2 + 7x + 10, (c) 2x2 - 5x - 3, (d) 15x2 + 19x + 6, (e) 9x2 + 12x + 4, (f) 4x2 - 9, (g) 5x2 + 9x, (h) 6x2 + 11x - 10.
5.2.1 Roots
If we have u x v = 0 then we must have either u = 0 or v = 0 or both u and v are 0. This is because 0 times any number is 0. Thus if we have the quadratic equation x2 + 5x + 6 = 0 and rewrite it as (x + 2)(x + 3) = 0, then we must have either x + 2 = 0, or x + 3 = 0 or both equal to 0. This means that the solutions to the quadratic equation are the solutions of these two linear equations, i.e. x = - 2 and x = - 3 . These values are called the roots of the equation. We can check these values by substituting them into the quadratic equation. Thus for x = - 2 we have 4 - 10 + 6 = 0 and thus 0 = 0.
For x = - 3 we have 9 - 15 + 6 = 0 and thus 0 = 0.
Thus we can solve a quadratic equation by the following:
1 Factorise the quadratic 2 Set each factor equal to 0
3 Solve the resulting linear equations Example
Factorise and hence solve the quadratic equation x2 - 3x + 2 = 0.
This equation is of the form given by equation [1] above. To factorise this equation we need to find the two numbers which when multiplied together will give 2 and which when added together will give - 3 . If we multiply - 1 and - 2 we obtain 2 and the addition of - 1 and - 2 gives - 3 . Thus we can write:
(x - l)(x - 2) = 0
The solutions are thus given by x - 1 = 0, i.e. x = 1, and x - 2 = 0, i.e.
x = 2.
We can check these values by substituting them into the original equation, x2 - 3x + 2. Thus, for x = 1 we have 1 - 3 + 2 = 0 and so 0 = 0. For x = 2 we have 4 - 6 + 2 = 0 and so 0 = 0.
Example
Factorise and hence solve the quadratic equation x2 - 4 = 0.
This equation is of the form given by equation [4] above. We can write this equation in the form x2 + 0x - 4 = 0. We thus need to find a product of two numbers which gives - 4 and a sum which gives 0.
Hence the factors are:
(x - 2)(x + 2) = 0
Thus the solutions are x - 2 = 0, i.e. x = 2, and x + 2 = 0, i.e. x = - 2 . These values can be checked by substituting them in the original equation, x2 - 4 = 0. With x = 2 we have 4 - 4 = 0 and with x = - 2 we have 4 - 4 = 0.
Example
Factorise and hence solve the quadratic equation x2 + 6x + 9 = 0.
This is of the form given by equation [2] above. The product of two numbers is 9 and the sum is 6. Hence
(x + 3)(x + 3) = 0
The solutions are thus given by x + 3 = 0, i.e. x = - 3 , and x + 3 = 0, i.e.
x = - 3 . There are two roots with the same value. We can check these values by substituting them in the original equation, x2 + 6x + 9 = 0.
Thus 9 - 18 + 9 = 0.
Example
Factorise and hence solve the quadratic equation x2 + 5x = 0.
This is of the form of equation [5] given above. We can rewrite this equation in the form x(x + 5) = 0. Hence the solutions are given by x = 0 or x + 5 = 0, i.e. x = - 5 . we can check these values by substituting them in the original equation, x2 + 5x = 0. Thus with x = 0 w e have 0 + 0 = 0 and with x = - 5 we have 25 - 25 = 0.
Example
If the roots of a quadratic equation are +2 and +1, what is the equation?
We have x = +2 and so x - 2 = 0, and x = +1 and so x - 1 = 0. Thus we can write the equation as
(x - 2)(x - 1) = 0
Multiplying this out gives x2 - 3x + 2 = 0.
Revision
3 Factorise and hence solve the following quadratic equations:
(a)x2 + x - 2 = 0, (b) x2 + 2x - 8 = 0, (c) x2 - 5x + 4 = 0,
(d) x2 - 4 = 0, (e) x2 + 4x + 4 = 0, (f) x2 - 2x + 1 = 0, (g) x2 + 7x = 0, (h) x2 - 1 = 0, (i) x2 + 2x - 3 = 0, (j) x2 - 5x + 6 = 0.
Determine the quadratic equations giving the following roots:
(a) x = 1, x = 4, (b) x = - 1 , x = 2, (c) x = 0, x = 6, (d) x = - 2 , x = - 3 , (e) x = 2, x = 0, (f) x = - 1 , x = - 2 , (g) x = 3, x = 1.
5.3 Completing the