Title: Default for 3.2 Solving Quadratic Equations
Course: HELM Workbooks 2023-2024

Quadratic equations

A generalisation of the third law of indices states:

Key Point

A quadratic equation is one which can be written in the form

\[ax^2+bx+c=0\qquad a\ne 0\]

where \(a\), \(b\) and \(c\) are given numbers and \(x\) is the unknown whose value(s) must be found.

For example

\[2x^2+7x-3=0, \qquad x^2+x+1=0, \qquad 0.5x^2+3x+9=0\]

are all quadratic equations. To ensure the presence of the \(x^2\) term, the number \(a\), in the general expression \(ax^2+bx+c\) cannot be zero.

However \(b\) or \(c\) may be zero, so that

\[4x^2+3x=0, \quad 2x^2-3=0 \quad \hbox{ and }\quad  6x^2=0\]

are also quadratic equations. Frequently, quadratic equations occur in non-standard form but where necessary they can be rearranged into standard form. For example

\[3x^2+5x=8, \qquad \hbox{ can be re-written as } \qquad  3x^2+5x-8=0;\]

\[2x^2=8x-9, \qquad \hbox{ can be re-written as } \quad 2x^2-8x+9=0;\]

\[1+x=\frac{1}{x}, \qquad\ \quad \hbox{ can be re-written as } \quad x^2+x-1=0.\]

A little care should be taken in the last example as in case one found \(x=0\) as a solution, this would not work in the initial equation. To solve a given quadratic equation we must find values of the unknown \(x\) which make the left-hand and right-hand sides equal. Such values are known as solutions or roots of the quadratic equation.

Note the difference between solving quadratic equations in comparison to solving linear equations. A quadratic equation will generally have two values of \(x\) (solutions) which satisfy it whereas a linear equation only has one solution.

We shall now describe three techniques for solving quadratic equations:

1. factorisation;
2. completing the square;
3. using the quadratic formula.

Solution by Factorisation

It may be possible to solve a quadratic equation by factorisation, essentially being able to spot the roots. You should be aware that not all quadratic equations can be easily factorised but some simple cases can occur.

Example

Solve the equation \(x^2+5x=0.\)

Solution

Factorising and equating each factor to zero we find

\[x^2+5x =0\quad \text{is equivalent to}\quad x(x+5) = 0\]

so that \(x=0\) and \(x=-5\) are the two solutions. We know this since the product of two numbers (the factors, in this case) will be zero if at least one of them is zero.

Example

Solve the quadratic equation \(x^2+x-6=0\).

Solution

Factorising the left hand side we find \(x^2+x-6=(x+3)(x-2)\) so that

\[x^2+x-6=0 \quad \text{is equivalent to}\quad (x+3)(x-2) = 0.\]

When the product of two quantities equals zero, at least one of the two must equal zero. In this case either \((x+3)\) is zero or \((x-2)\) is zero. It follows that

\[x+3=0, \qquad \hbox{ giving } \qquad x = -3\]

or

\[x-2 = 0,\qquad  \hbox{ giving } \qquad x=2\]

Here there are two solutions, \(x=-3\) and \(x=2\). These solutions can be checked quite easily by substitution back into the given equation.

Example

Solve the quadratic equation \(2x^2-7x-4=0\) by factorising the left-hand side.

Solution

Factorising the left hand side: \(2x^2-7x-4=(2x+1)(x-4)\) so \(2x^2-7x-4=0\) \(\text{is equivalent to}\ \  (2x+1)(x-4) = 0.\)

In this case either \((2x+1)\) is zero or \((x-4)\) is zero. It follows that \(2x+1=0, \quad \text{giving } \quad x = -\dfrac{1}{2}\quad\) or \( x-4 = 0,\quad  \text{giving } \quad x=4\).

There are two solutions, \(x=-\dfrac{1}{2}\) and \(x=4\).

Example

Solve the equation \(4x^2+12x+9=0\).

Solution

Factorising we find \(4x^2+12x+9 = (2x+3)(2x+3) = (2x+3)^2\). This time the factor \((2x+3)\) occurs twice. The original equation \(4x^2+12x+9=0\) becomes

\[(2x+3)^2=0\]

so that \(2x+3 = 0\) and we obtain the solution \(x=-\dfrac{3}{2}\). Because the factor \(2x+3\) appears twice in the equation \((2x+3)^2=0\) we say that this root is a repeated solution or double root.

Exercise

Solve the quadratic equation \([value]=0\). If there is a repeated root, enter this twice.

\(x = \) or \(x = \)
Feedback: We are asked to factorise the quadratic equation \([value]=0.\) First factorise the left-hand side: \[[value]=[value].\] Equate each factor to zero to obtain the two solutions: Solution 1: \(x= [value]\) Solution 2: \(x= [value]\).

Exercise

Solve the quadratic equation \([value]=0\). If there is a repeated root, enter this twice.

\(x = \) or \(x = \)
Feedback: We are asked to factorise the quadratic equation \([value]=0.\) First factorise the left-hand side: \[[value]=[value].\] Equate each factor to zero to obtain the two solutions: Solution 1: \(x= [value].\) Solution 2: \(x= [value].\)

Completing the square

The technique known as completing the square can be used to solve quadratic equations although it is applicable in many other circumstances too so it is well worth studying.

Observe that the quadratic \(x^2+c=0\), is more easily solved than \(x^2+bx+c=0\). The method of completing the square involves rewriting the equation so that the variable \(x\) appears only in the square term. That is, we can write \( (x+d)^2+e=0\) for suitable numbers \(d,e\). This form is not just easier to solve, it reveals more information about the graph of \(y=x^2+ax+c\), and will be used to simplify expressions appearing in later material such as in calculus.

Example

1. Show that \((x+3)^2=x^2+6x+9\).
2. Hence show that \(x^2+6x\) can be written as \((x+3)^2 - 9\).

Solution

1. Removing the brackets we find \[(x+3)^2= (x+3)(x+3) = x^2 +3x+3x + 9 = x^2 +6x + 9.\]
2. By subtracting \(9\) from both sides of the previous equation it follows that \[(x+3)^2-9 = x^2+6x.\]

Example

1. Show that \((x-4)^2= x^2-8x+16\).
2. Hence show that \(x^2-8x\) can be written as \((x-4)^2-16\).

Solution

1. Removing the brackets we find \[(x-4)^2= (x-4)(x-4) = x^2-4x-4x+16 = x^2-8x+16.\]
2. Subtracting \(16\) from both sides we can write \[(x-4)^2-16 = x^2-8x.\]

We shall now generalise the results of the previous two examples. Noting that

\[(x+k)^2=x^2+2kx+k^2\]

we can write

\[ x^2+2kx = (x+k)^2 - k^2.\]

Note that the constant term in the brackets on the right-hand side is always half the coefficient of \(x\) on the left. This process is called completing the square.

Key Point

Completing the Square

\[\qquad \hbox{The expression}\quad x^2+2kx \quad\hbox{is equivalent to}\quad (x+k)^2 -k^2.\]

Example

Complete the square for the expression \(x^2+16x\).

Solution

Comparing \(x^2+16x\) with the general form \(x^2+2kx\) we see that \(k=8\). Hence

\[x^2+16x = (x+8)^2 - 8^2 = (x+8)^2-64\]

Note that the constant term in the brackets on the right, that is \(8\), is half the coefficient of \(x\) on the left, which is \(16\).

Example

Complete the square for the expression \(5x^2+4x\).

Solution

Consider \(5x^2+4x\). First of all the coefficient \(5\) is removed outside a bracket as follows

\[ 5x^2+4x = 5\left(x^2+{4\over 5}x\right).\]

We can now complete the square for the quadratic expression in the brackets:

\[ x^2+{4\over 5}x = \left(x+{2\over 5}\right)^2 - \left({2\over 5}\right)^2 = \left(x+{2\over 5}\right)^2-{4\over25}.\]

Finally, multiplying both sides by \(5\) we find

\[ 5x^2 + 4x = 5\left(\left(x+{2\over 5}\right)^2- {4\over 25}\right).\]

Completing the square can be used to solve quadratic equations as shown in the following examples.

Example

Solve the equation \(x^2+6x+2=0\) by completing the square.

Solution

First of all just consider \(x^2+6x\), and note that we can write this as \[x^2+6x = (x+3)^2 - 9.\]

Then the quadratic equation can be written as

\[x^2+6x+2= (x+3)^2-9 + 2  =0\]

that is

\[ (x+3)^2 = 7.\]

Taking the square root of both sides gives

\[x+3 = \pm \sqrt 7\]

so \(x = -3 \pm \sqrt 7\). The two solutions are \(x=-3+\sqrt{7}=-0.3542\) and \(x=-3-\sqrt{7}=-5.6458\), to 4 d.p.

Example

Solve the equation \(x^2-8x+5=0\).

Solution

First consider \(x^2-8x\) which we can write as

\[x^2-8x  = (x-4)^2 -16\]

so that the equation becomes

\[x^2-8x+5 = (x-4)^2-16 + 5 = 0.\]

i.e \[(x-4)^2 = 11 \text{ implies } x-4 = \pm\sqrt{11} \text{ and hence } x= 4 \pm \sqrt{11}.\]

So \(x=7.3166\) or \(x= 0.6834\) (to 4 d.p.).

Exercise

Solve the equation \(x^2-[value]x+[value]=0\) by completing the square.

\(x =\) or \(x =\)
Feedback: First examine the two left-most terms in the equation: \(x^2-[value]x\). Complete the square for these terms: \[x^2-[value]x = \left(x-\frac{[value]}{2}\right)^2- \left(\frac{[value]}{2}\right)^2\] Use the above result to rewrite the equation \(x^2-[value]x+[value]=0\) in the form \((x+d)^2+e=0\): \[x^2-[value]x+[value]= \left(x-\frac{[value]}{2}\right)^2-\frac{[value]^2}{4}+[value] = [value]=0.\] From this now obtain the roots: \[[value]=0, \text{ so }\ x= [value] \text{ or } x= [value].\]

Exercise

Solve the equation \(4x^2-[value]x+[value]=0\) by completing the square.

\(x =\) or \(x =\)
Feedback: First examine the two left-most terms in the equation: \(4x^2-[value]x\). Complete the square for these terms: \[4x^2-[value]x = 4\left(x^2-\frac{[value]}{4} x\right)=4\left(x-\frac{[value]}{8}\right)^2-\frac{a^2}{16}.\] Use the above result to rewrite the equation \(4x^2-[value]x+[value]\) in the form \(2(x-\ ?)^2+?=0\): \[4x^2-[value]x+[value] = 4\left(x-\frac{[value]}{8}\right)^2-\frac{[value]^2}{16}+1 =0.\] From this now obtain the roots: \[\left(x-\frac{[value]}{8}\right)^2=\frac{1}{4}\left(\frac{[value]^2}{16}-1\right). \text{Therefore}\quad x=[value]/8\pm \frac{\sqrt{[value]^2/16-[value]}}{2}.\]

Exercise

A chemical manufacturer found that the sales figures for a certain chemical X\(_2\)O depended on its selling price. At present, the company can sell all of its weekly production of \(300\text{kg}\) at a price of £\(600\) per \(\text{kg}\).  The company's market research department advised that the amount sold would decrease by only \(1\) \(\text{kg}\) per week for every £\(2\) per \(\text{kg}\) increase in the price of X\(_2\)O. If the total production costs are made up of a fixed cost of £\(30000\) per week, plus £\(400\) per \(\text{kg}\) of product.

Let £\(x\) be the new price per \(\text{kg}\) of X\(_2\)O. Find a quadratic polynomial \(P\) in \(x\) which gives the weekly profit.

\(P =\)

Complete the square for the above  expression and hence find

1. the price which maximises the weekly profit on sales of X\(_2\)O;  £
2. the maximum weekly profit;  £
3. the weekly production rate.
Feedback: Firstly, we must derive the equation for weekly profit. If \(x\) is the new price per kg of the chemical, then the new amount produced is \[300-\frac{(x-600)}{2} = 600-\frac{x}{2}.\] Now we must work out the profit. This will be what we make from selling the product minus any costs. \[P = x\left(600-\frac{x}{2}\right)-30000-400\left(600-\frac{x}{2}\right) =\] \[-\dfrac{x^2}{2}+800x-270000.\] Completing the square we obtain \[P = -\frac{1}{2}(x^2-1600x+540000) = -\frac{1}{2}((x-800)^2+540000-800^2 =\] \[ -\frac{1}{2}((x-800)^2+540000-800^2) =-\frac{1}{2}((x-800)^2-100000).\] 1. We want to maximise \(P\). This will occur when \((x-800)^2=0\), so \(x=800\). 2. At \(x=800\), \(P = 50000\). 3. The production rate is \(600-\frac{x}{2}\). Subbing in our answer from part 1 we obtain \[600-\frac{800}{2} = 200.\]

Solution by formula

When it is difficult to factorise a quadratic equation, it may be possible to solve it using a formula which is used to calculate the roots. The formula is obtained by completing the square in the general quadratic \(ax^2+bx+c=0\). We proceed by removing the coefficient of \(a\):

\[ax^2+bx+c= a\left(x^2+\frac{b}{a}x+\frac{c}{a}\right)=a\left(\left(x+\frac{b}{2a}\right)^2+\frac{c}{a}-\frac{b^2} {4a^2}\right).\]

Thus the solution of \(ax^2+bx+c=0\) is the same as the solution to

\[\left(x+\frac{b}{2a}\right)^2+\frac{c}{a}-\frac{b^2}{4a^2}=0.\]

So, solving:

\[\left(x+\dfrac{b}{2a}\right)^2=-\dfrac{c}{a}+\dfrac{b^2}{4a^2}\qquad\text{which leads to}\qquad x=-\dfrac{b}{2a}\pm\sqrt{ -\dfrac{c}{a}+\dfrac{b^2}{4a^2}}.\]

Simplifying this expression further we obtain the important result:

A generalisation of the third law of indices states:

Key Point

Quadratic Formula

If \(ax^2+bx+c=0,\ \ a\ne 0\) then the two solutions (roots) are

\[x= {-b- \sqrt{b^2-4ac}\over 2a}\qquad\text{and}\qquad x= {-b+ \sqrt{b^2-4ac}\over 2a}.\]

To apply the formula to a specific quadratic equation it is necessary to identify carefully the values of \(a\), \(b\) and \(c\), paying particular attention to the signs of these numbers. Substitution of these values into the formula then gives the desired solutions. Note that if the quantity \(b^2-4ac\) (called the discriminant) is a positive number we can take its square root and the formula will produce two values known as distinct real roots. If \(b^2-4ac=0\) there will be one value only known as a repeated root or double root. The value of this root is \(x=-\dfrac{b}{2a}\). Finally, if \(b^2-4ac\) is negative we say the equation possesses complex roots. These require special treatment.

A generalisation of the third law of indices states:

Key Point

When finding roots of the quadratic equation \(ax^2+bx+c=0\) first calculate the discriminant

\(b^2-4ac.\)

- If \(b^2-4ac>0\)  the quadratic has two real distinct roots.
- If \(b^2-4ac=0\) the quadratic has two real and equal roots.
- If \(b^2-4ac<0\) the quadratic has no real roots. There are two complex roots.

Example

Compare each given equation with the standard form \(ax^2+bx+c=0\) and identify \(a\), \(b\) and \(c\). Calculate \(b^2-4ac\) in each case and use this information to state the nature of the roots.

1. \(3x^2+2x-7=0\)
2. \(3x^2+2x+7=0\)
3. \(3x^2-2x+7=0\)
4. \(x^2+x+2=0\)
5. \(-x^2+3x-{1\over 2}=0\)
6. \(5x^2-3=0\)
7. \(x^2-2x+1=0\)
8. \(2p^2-4p=0\)
9. \(-p^2+4p-4=0\)

Solution

1. \(a=3\), \(b=2\), \(c=-7\). So \(b^2-4ac=(2)^2-4(3)(-7)=88\). The roots are real and distinct.
2. \(a=3\), \(b=2\),\(c=7\). So \(b^2-4ac=(2)^2-4(3)(7)=-80\). The roots are complex.
3. \(a=3\), \(b=-2\), \(c=7\). So \(b^2-4ac=(-2)^2-4(3)(7)=-80\). The roots are complex.
4. \(a=1\), \(b=1\), \(c=2\). So \(b^2-4ac=1^2-4(1)(2)=-7\). The roots are complex.
5. \(a=-1\), \(b=3\), \(c=-{1\over 2}\). So \(b^2-4ac=3^2-4(-1)(-{1\over 2})=7\). The roots are real and distinct.
6. \(a=5\), \(b=0\), \(c=-3\). So \(b^2-4ac=0-4(5)(-3)=60\). The roots are real and distinct.
7. \(a=1\), \(b=-2\), \(c=1\). So \(b^2-4ac=(-2)^2-4(1)(1)=0\). The roots are real and equal.
8. \(a=2, b=-4, c=0\). So \(b^2-4ac=(-4)^2-4(2)(0)=16\). The roots are real and distinct.
9. \(a=-1, b=4, c=-4\). So \(b^2-4ac=(-4)^2-4(-1)(-4)=0\). The roots are real and equal.

Example

Solve the quadratic equation \(2x^2+3x-6=0\) using the formula.

Solution

We compare the given equation with the standard form \(ax^2+bx+c=0\) in order to identify \(a\), \(b\) and \(c\). We see that here \(a=2\), \(b=3\) and \(c=-6\). Note particularly the sign of \(c\). Substituting these values into the formula we find

\[x =  \frac{-b\pm \sqrt{b^2-4ac}}{ 2a} = \frac{-3 \pm \sqrt{3^2 - 4(2)(-6)}}{(2)(2)}=\frac{-3 \pm \sqrt{9+48}}{4} = \frac{-3 \pm 7.5498}{4} .\]

Hence, to 4 d.p., the two roots are \(x=1.1375\), if the positive sign is taken and \(x=-2.6375\) if the negative sign is taken. However, it is often sufficient to leave the solution in the so-called surd form \(x=\dfrac{-3\pm\sqrt{57}}{4}\), which is exact.

Exercise

Solve the equation \([value]=0\) using the quadratic formula. Give your answer in the form of a set.
Feedback: We are given the equation the equation \([value]=0.\) First identify \(a\), \(b\) and \(c\): \(a=[value]\), \(b=[value]\), \(c=[value]\). Substitute these values into the formula and simplify: \[x = \frac{-b\pm \sqrt{b^2-4ac}}{2a} = \dfrac{-([value])\pm\sqrt{[value]^2-(4)([value])([value])}}{(2)([value])}.\] Hence \([value]\), though the answer is fine without the "\(x=\)".

Engineering Example 1

Undersea cable fault location

Introduction

The voltage (\(V\)), current (\(I\)) and resistance (\(R\)) in an electrical circuit are related by Ohm's law i.e. \(V = IR\). If there are two resistances \((R_1\) and \( R_2)\) in an electrical circuit, they may be in series, in which case the total resistance (\(R\)) is given by \(R = R_1 + R_2\). Or they may be in parallel in which case the total resistance is given by

\[\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}.\]

In 1871 the telephone cable between England (\(A\)) and Denmark (\(B\)) developed a fault, due to a short circuit under the sea (see the figure below). Oliver Heaviside, an electrical engineer, came up with a very simple method to find the location of the fault. He assumed that the cable had a uniform resistance per unit length. Heaviside performed two tests:

1. connecting a battery (voltage \(E\)) at \(A\), with the circuit open at \(B\), he measured the resulting current \(I_1\);
2. connecting the same battery at \(A\), with the cable earthed at \(B\), he measured the current \(I_2\).

[Image: 3.2 Schematic of the Undersea cable]

 Schematic of the undersea cable

In the first measurement the resistances up to the cable fault and between the fault and the short circuit are in series and in the second experiment the resistances beyond the fault and between the fault and the short circuit are in parallel.

Problem in words

Use the information from the measurements to deduce the location of the fault.

Mathematical statement of problem

1. Denote the resistances of the various branches by the symbols shown in the figure above.
2. Use Ohm's law to write down expressions that apply to each of the two measurements.
3. Eliminate \(y\) from these expressions to obtain an expression for \(x\).

Mathematical analysis

1. In the first experiment the total circuit resistance is \(x + y\). In the second experiment, the total circuit resistance is given by: \[\qquad x+\left(\dfrac{1}{r-x}+\dfrac{1}{y}\right)^{-1}\]
2. Application of Ohm's law to each experimental situation gives:\[\qquad E=I_1(x+y)\quad (1)\] \[\qquad E=I_2\left(x+\left(\dfrac{1}{r-x}+\dfrac{1}{y}\right)^{-1} \right)\quad (2)\]
3. Rearrange Equation (1) to give \[\dfrac{E}{I_1}-x=y.\]
Substitute for \(y\) in Equation (2), divide both sides by \(I_2\) and introduce \(\dfrac{E}{I_1}=r_1\)   and  \(\dfrac{E}{I_2}=r_2\):
\[\qquad r_2=\left(x+\left(\dfrac{1}{r-x}+\dfrac{1}{r_1 - x}\right)^{-1}\right).\]
Use a common denominator for the fractions on the right-hand side:
\[\qquad r_2=\left(x+\left(\dfrac{(r-x)(r_1-x)}{r_1-x+r-x}\right)\right)=\dfrac{x(r_1+r-2x)+(r-x)(r_1-x)}{(r_1+r-2x)}.\]
Multiply through by \((r_1+r-2x)\):
\[\qquad r_2(r_1+r-2x)=x(r_1+r-2x)+(r-x)(r_1-x).\]
Rearrange as a quadratic for \(x\):
\[\qquad x^2-2r_2x-rr_1+r_2r_1+rr_2=0.\]
Use the standard formula for solving quadratic equations with \(a = 1, b = -2r_2\) and \(c=-rr_1+r_2r_1+rr_2\):
\[x=\dfrac{2r_2\pm\sqrt{4r_2^2-4(-rr_1+r_2r_1+rr_2)}}{2}=r_2\pm\sqrt{(r-r_2)(r_1-r_2)}.\]
Only positive solutions would be of interest.

Engineering Example 2

Estimating the mass of a pipe

Introduction

Sometimes engineers have to estimate component weights from dimensions and material properties. On some occasions, engineers prefer use of approximate formulae to exact ones as long as they are sufficiently accurate for the purpose. This example introduces both of these aspects.

Problem in words

1. Find the mass of a given length of pipe in terms of its inner and outer diameters and the density of the pipe material.
2. Find the wall thickness of the pipe if the inner diameter is \(0.15 \; \textrm{m}\), the density is \(7900 \; \textrm{kg m}^{-3}\) and the mass per unit length of pipe is \(40 \; \textrm{kg m}^{-1}\).
3. Find an approximate method for calculating the mass of a given length of a thin-walled pipe and calculate the maximum ratio of inner and outer diameters that give an error of less than \(10\) % when using the approximate method.

Mathematical statement of problem

1. Denote the length of the pipe by \(L \; \textrm{m}\) and inside and outside diameters by \(d_i \; \textrm{m}\) and \(d_o \; \textrm{m}\), respectively and the density by \(\rho \; \textrm{kg m}^{-3}\).  Assume that the pipe is cylindrical so its cross section corresponds to the gap between concentric circles (this is called an annulus or annular region). Calculate the difference in cross sectional areas by using the formula for the area of a circle (\(A = \pi r^2\) where \(r\) is the radius) and multiply by the density and length to obtain mass (\(m\)).
2. Rearrange the equation in terms of wall thickness (\(d \; \textrm{m}\)) and inner diameter. Substitute the given values to determine the wall thickness.
3. Approximate the resulting expression for small values of (\(d_o-d_i\)). Calculate the percentage difference in predictions between the original and approximate formulae for various numerical values of \(d_i / d_o\).

Mathematical analysis

1. The cross section of a cylindrical pipe is a circular annulus. The area of a circle is given by \(\pi r^2=\dfrac{\pi}{4}d^2\), since \(r = d/2\) if \(d\) is the diameter.  So the area of the outer circle is \(\dfrac{\pi}{4}d_o^2\)  and that of the inner circle is \(\dfrac{\pi}{4}d_i^2\).  This means that the mass \(m \; \textrm{kg}\) of length \(L \; \textrm{m}\) of the pipe is given by \[m=\dfrac{\pi}{4}(d_0^2-d_i^2)L\rho.\]
2. Denote the pipe wall thickness by \(\delta\) so \(d_o=d_i+2\delta\). Use \((d_o^2-d_i^2)=(d_o-d_i)(d_o+d_i)=2\delta (2d_i+2\delta)\). So \[\quad m=\pi\delta (d_i +\delta)L\rho\]
Given that \(m/L = 40,\ d_i = 0.15\) and \( \rho = 7900\),
\[\qquad 40 = \pi \delta(0.15 + \delta)7900.\]
Rearrange this equation as a quadratic in \(\delta\),
\[\qquad \delta^2+0.15\delta -4/(790\pi) =0.\]
Solve this quadratic using the standard formula with \(a = 1, b = 0.15\) and \(c = -4/(790\pi)\) to find solutions
\[ \frac{-0.15 \pm \sqrt{(0.15)^2-(4)(1)(-4/(790\pi))}}{2}\] Retain only the positive solution to give \(\delta = 0.01\) to 2 decimal places, i.e. the pipe wall thickness is \(10 \; \textrm{mm}\).
3. If \(\delta\) is small then \((d_o - d_i)\) is small and \(d_i + \delta \approx d_i\). So the expression for \(m\) in terms of \(\delta\) may be written
\[ m\approx \pi\delta d_i L\rho.\]
The graph below shows that the percentage error from using the approximate formula for the mass of the pipe exceeds \(10\)%  only if the inner diameter is less than \(82\)% of the outer diameter.

The percentage error from using the approximate formula can be calculated from (exact result \(-\) approximate result)/(exact result) \(\times\) \(100\)% for various values of the ratio of inner to outer diameters. In the graph the error is plotted for diameter ratios between \(0.75\) and \(1\).

[Image: 3.2 Engineering Example 2]

Percentage Error for different diameters

Comment

The graph shows also that the error is \(1\)% or less for diameter ratios \(> 0.98\).

Geometrical representation of quadratics

We can plot a graph of the function \(y=ax^2+bx+c\) (given the values of \(a,b\) and \(c\)). If the graph crosses the horizontal axis it will do so when \(y=0\), and so the \(x\) coordinates at such points are solutions of \(ax^2+bx+c=0\). Depending on the sign of \(a\) and of the nature of the solutions there are essentially six different types of graph that can occur. These are displayed in the figure below.

[Image: 3.2 The possible graphs of a quadratic y = ax^2 + bx + c]

The possible graphs of a quadratic \(y=ax^2+bx+c\).

Sometimes a graph of the quadratic is used to locate the solutions; however, this approach is generally inaccurate.   This is illustrated in the following example.

Example

Solve the equation \(x^2-4x+1=0\) by plotting a graph of the function: \[ y=x^2-4x+1.\]

Solution

By constructing a table of function values we can plot the graph as shown in the figure below

[Image: The graph of y = x^2 − 4x + 1 cuts the x axis at C and D]

The graph of \(y=x^2-4x+1\) cuts the \(x\) axis at \(C\) and \(D\)

The solutions of the equation \(x^2-4x+1=0\) are found by looking for points where the graph crosses the horizontal axis. The two points are approximately \(x=0.3\) and \(x=3.7\) marked \(C\) and \(D\) on the figure.

Solve the equation \([value]=0\) using the quadratic formula.

\(x_1 =\)

\(x_2 =\)
Feedback: First identify \(a\), \(b\) and \(c\): \(a=[value],\qquad b=[value], \qquad c=[value]\). Substitute these values into the formula and simplify: \[x = \frac{-b\pm \sqrt{b^2-4ac}}{2a} = \dfrac{-[value]\pm\sqrt{[value]^2-(4)([value])([value])}}{(2)([value])}.\] Hence \([value]\).

Exercise

Solve the quadratic equation \([value]=0\). If there is a repeated root, enter this twice.

\(x = \) or \(x = \)
Feedback: We are given the quadratic equation \([value]=0.\) First factorise the left-hand side: \[[value]=[value].\] Equate each factor is then equated to zero to obtain the two solutions: Solution 1: \(x= [value]\) Solution 2: \(x= [value]\).

Exercise

Solve the equation \([value]=0\) using the quadratic formula. Give your answer in the form of a set.
Feedback: First identify \(a\), \(b\) and \(c\): \(a=[value]\), \(b=[value]\), \(c=[value]\). Substitute these values into the formula and simplify:\[x = \frac{-b\pm \sqrt{b^2-4ac}}{2a} = \dfrac{ [value] \pm \sqrt{ ([value])^2-4([value])([value])} } {(2)([value])}.\] Hence \([value]\).

Exercise

Solve the equation \(4x^2-[value]x+[value]=0\) using the quadratic formula.

\(x =\) or \(x =\)
Feedback: First identify \(a\), \(b\) and \(c\): \(a=4\), \(b=-[value]\), \(c=[value]\). Substitute these values into the formula and simplify: \[x = \frac{-b\pm \sqrt{b^2-4ac}}{2a} = \dfrac{[value]\pm\sqrt{[value]^2-(4)(4)([value])}}{(2)(4)} = \dfrac{[value]\pm\sqrt{[value]-[value]}}{8}.\] Hence \(x = [value]\) and \(x = [value]\).