Roots of Quadratic Equations (Edexcel IGCSE Further Pure Maths)

Revision Note

Roger B

Written by: Roger B

Reviewed by: Dan Finlay

Discriminants

What is the discriminant of a quadratic function?

  • The discriminant of a quadratic is often denoted by the Greek letter straight capital delta (upper case delta)

  • For a quadratic  a x squared plus b x plus c space space open parentheses a not equal to 0 close parentheses the discriminant is given by

    • straight capital delta equals b squared minus 4 a c 

  • The discriminant is the expression that is inside the square root in the quadratic formula

    • x equals fraction numerator negative b plus-or-minus square root of b squared minus 4 a c end root over denominator 2 a end fraction

  • This is not on the exam formula sheet so you need to remember it

How does the discriminant of a quadratic function affect its graph and roots?

  • The discriminant tells us about the roots (or solutions) of the equation  a x squared plus b x plus c equals 0

    • It also tells us about the graph of  y equals a x squared plus b x plus c

  • If Δ > 0 then square root of b squared minus 4 a c end root and negative square root of b squared minus 4 a c end root are two distinct values

    • The equation a x squared plus b x plus c equals 0 has unequal real roots

      • i.e. there are two distinct real solutions

    • The graph of space y equals a x squared plus b x plus c crosses the x-axis twice

  • If straight capital delta equals 0 then square root of b squared minus 4 a c end root and negative square root of b squared minus 4 a c end root are both zero

    • The equation a x squared plus b x plus c equals 0 has equal real roots

      • i.e. it has one repeated real solution

    • The graph of space y equals a x squared plus b x plus c touches the x-axis at exactly one point

      • This means that the x-axis is a tangent to the graph

  • If straight capital delta less than 0 then square root of b squared minus 4 a c end root and negative square root of b squared minus 4 a c end root are both undefined

    •   The roots of the equation a x squared plus b x plus c equals 0 are not real

      • i.e. it has no real solutions

    • The graph of space y equals a x squared plus b x plus c never touches the x-axis

      • This means that graph is wholly above (or below) the x-axis

Discrimamts Notes Diagram 2

How do I solve problems using the discriminant?

  • Often at least one of the coefficients of a quadratic will be given as an unknown

    • For example the letter k may be used for the unknown constant

  • You will be given a fact about the quadratic such as:

    • The number of real solutions of the equation

    • The number of roots (i.e. x-intercepts) of the graph

  • To find the value or range of values of k

    • Find an expression for the discriminant

      • Use straight capital delta equals b squared minus 4 a c

    • Decide whether straight capital delta greater than 0, straight capital delta equals 0 or straight capital delta less than 0

      • If the question says there are real roots but does not specify how many then use straight capital delta greater or equal than 0

    • Solve the resulting equation or inequality for k

Examiner Tips and Tricks

  • Questions won't always use the word discriminant

    • It is important to recognise when its use is required

    • Look for

      • a number of roots or solutions being stated

      • whether and/or how often the graph of a quadratic function intercepts the x-axis

Worked Example

A function is given by space straight f left parenthesis x right parenthesis equals 2 k x squared plus k x minus k plus 2 , where k is a constant. The graph of space y equals f left parenthesis x right parenthesis intercepts the x-axis at two different points.

a) Show that 9 k squared minus 16 k greater than 0.

The question says the graph 'intercepts the x-axis at two different points'

This means that the discriminant b squared minus 4 a c is greater than zero

Here  a equals 2 k,  b equals k,  and  c equals negative k plus 2

open parentheses k close parentheses squared minus 4 open parentheses 2 k close parentheses open parentheses negative k plus 2 close parentheses greater than 0

Expand the brackets and collect terms

table row cell k squared minus 8 k open parentheses negative k plus 2 close parentheses end cell greater than 0 row cell k squared plus 8 k squared minus 16 k end cell greater than 0 end table

bold 9 bold italic k to the power of bold 2 bold minus bold 16 bold italic k bold greater than bold 0

b) Hence find the range of possible values of k.

Solve the inequality, beginning by factorising

k open parentheses 9 k minus 16 close parentheses greater than 0

This tells us the graph of  y equals 9 k squared minus 16 k  intercepts the horizontal axis at k equals 0 and k equals 16 over 9

It can be helpful to sketch a graph here

Graph for solving quadratic inequality


9 k squared minus 16 k greater than 0 will be true to the left of 0 and to the right of 16 over 9

Write these down as inequalities

bold italic k bold less than bold 0 bold space bold space bold or bold space bold space bold italic k bold greater than bold 16 over bold 9

Sum & Product of Roots

How are the roots of a quadratic equation linked to its coefficients?

  • A quadratic equation a x squared plus b x plus c equals 0 (where a not equal to 0) has roots alpha and beta given by

    • space alpha comma beta equals fraction numerator negative b plus-or-minus square root of b squared minus 4 a c end root over denominator 2 a end fraction

      • i.e. the solutions found by the quadratic formula (or any other solution method)

  • This means the equation can be rewritten in the form  a left parenthesis x minus alpha right parenthesis left parenthesis x minus beta right parenthesis equals 0

    • Note that left parenthesis x minus alpha right parenthesis left parenthesis x minus beta right parenthesis equals x squared minus left parenthesis alpha plus beta right parenthesis x plus alpha beta

      • It is possible that the roots are repeated, i.e. that alpha equals beta

    • You can then equate the two forms:

      • a x squared plus b x plus c equals a left parenthesis x minus alpha right parenthesis left parenthesis x minus beta right parenthesis

    • Then (because a not equal to 0) you can divide both sides of that by a and expand the brackets:

      • x squared plus b over a x plus c over a equals x squared minus left parenthesis alpha plus beta right parenthesis x plus alpha beta

    • Finally, compare the coefficients

      • x coefficients:  negative b over a equals left parenthesis alpha plus beta right parenthesis

      • Constant terms:  c over a equals space alpha beta

  • Therefore for a quadratic equation a x squared plus b x plus c equals 0  :

    • The sum of the roots alpha plus beta is equal to negative b over a

    • The product of the roots alpha beta is equal to c over a

  • You don't need to prove these results on the exam

    • But you need to be able to use them to answer questions about quadratics

    • They are not on the exam formula sheet

      • So you need to remember (or be able to derive) them

  • You can use them

    • to find the sum and product of the roots if you know the equation

      • Just substitute ab and c into the formulae

    • or to find the equation if you know the sum and product of the roots

      • See the next section

How do I find a quadratic equation from information about its roots?

  • You may be given the sum and product of an equation's roots and then asked to find the equation

    • Usually the equation will need to have integer coefficients

      • For example  "A quadratic equation has roots alpha and beta, where  alpha plus beta equals 7 over 2  and  alpha beta equals negative 2. Find a quadratic equation with integer coefficients that has roots alpha and beta."

  • STEP 1
    Start with the formulae linking the roots and coefficients

    • alpha plus beta equals negative b over a 

      • So  b over a equals negative 7 over 2

    • alpha beta equals c over a

      • So  c over a equals negative 2

  • STEP 2
    Choose a value for a, and find the corresponding values for b and c 

    • Choose a value for a that will multiply to make the values for alpha plus beta and alpha beta into integers

      • Let a equals 2

      • Then  b equals negative 7 over 2 cross times 2 equals negative 7

      • And  c equals negative 2 cross times 2 equals negative 4

  • STEP 3
    Write down the equation using your values for ab and c

    • 2 x squared minus 7 x minus 4 equals 0

  • Note that the answer is not unique

    • Any multiple of the equation will also have the same roots

      • e.g.  4 x squared minus 14 x minus 8 equals 0

  • You may be asked to find a quadratic equation whose roots are related to the roots of another quadratic equation

    • I.e. whose roots are expressed in terms of the roots of the first equation

  • You will often need algebraic tricks to write other expressions with alpha and beta in terms of alpha plus beta and alpha beta

    • alpha and beta are the roots of the first quadratic

    • For example:

      • table row cell alpha squared plus beta squared end cell equals cell alpha squared plus 2 alpha beta plus beta squared minus 2 alpha beta end cell row blank equals cell open parentheses alpha plus beta close parentheses squared minus 2 alpha beta end cell end table

      • table row cell alpha cubed plus beta cubed end cell equals cell alpha cubed plus 3 alpha beta open parentheses alpha plus beta close parentheses plus beta cubed minus 3 alpha beta open parentheses alpha plus beta close parentheses end cell row blank equals cell alpha cubed plus 3 alpha squared beta plus 3 alpha beta squared plus beta cubed minus 3 alpha beta open parentheses alpha plus beta close parentheses end cell row blank equals cell open parentheses alpha plus beta close parentheses cubed minus 3 alpha beta open parentheses alpha plus beta close parentheses end cell end table

      • table row cell open parentheses alpha minus beta close parentheses squared end cell equals cell alpha squared minus 2 alpha beta plus beta squared end cell row blank equals cell alpha squared plus 2 alpha beta plus beta squared minus 4 alpha beta end cell row blank equals cell open parentheses alpha plus beta close parentheses squared minus 4 alpha beta end cell end table

  • Then if you know the values of alpha plus beta and alpha beta from the first quadratic, you can use them to find the sum or product of the new roots

    • For example "A quadratic equation has roots alpha and beta where  alpha plus beta equals negative 3 over 2  and  alpha beta equals 2.  Form a second quadratic equation with integer coefficients that has roots alpha over beta squared and beta over alpha squared."

    • The sum of the new roots is  alpha over beta squared plus beta over alpha squared equals fraction numerator alpha cubed plus beta cubed over denominator alpha squared beta squared end fraction equals fraction numerator alpha cubed plus beta cubed over denominator open parentheses alpha beta close parentheses squared end fraction

      • We can use the substitution for alpha cubed plus beta cubed from above

      • So  Sum equals fraction numerator open parentheses alpha plus beta close parentheses cubed minus 3 alpha beta open parentheses alpha plus beta close parentheses over denominator open parentheses alpha beta close parentheses squared end fraction equals fraction numerator open parentheses negative begin display style 3 over 2 end style close parentheses cubed minus 3 open parentheses 2 close parentheses open parentheses negative begin display style 3 over 2 end style close parentheses over denominator open parentheses 2 close parentheses squared end fraction equals 45 over 32

    • The product of the new roots is  alpha over beta squared cross times beta over alpha squared equals fraction numerator alpha beta over denominator alpha squared beta squared end fraction equals fraction numerator 1 over denominator alpha beta end fraction

      • So  Product equals fraction numerator 1 over denominator alpha beta end fraction equals 1 half

    • With the sum and product we can form the equation as described in the last section

      • 32 x squared minus 45 x plus 16 equals 0

Worked Example

The roots of the quadratic equation 2 x squared minus 11 x plus 5 equals 0 are alpha and beta.

Given that  alpha greater than beta  and without solving the equation,

(a) show that  alpha minus beta equals 9 over 2

Use  alpha plus beta equals negative b over a  and alpha beta equals c over a  to find the sum and product

alpha plus beta equals negative fraction numerator open parentheses negative 11 close parentheses over denominator 2 end fraction equals 11 over 2

alpha beta equals 5 over 2

Express open parentheses alpha minus beta close parentheses squared in terms of  alpha plus beta  and  alpha beta

table row cell open parentheses alpha minus beta close parentheses squared end cell equals cell alpha squared minus 2 alpha beta plus beta squared end cell row blank equals cell open parentheses alpha squared plus 2 alpha beta plus beta squared close parentheses minus 4 alpha beta end cell row blank equals cell open parentheses alpha plus beta close parentheses squared minus 4 alpha beta end cell end table

Substitute to find the value

table row cell open parentheses alpha minus beta close parentheses squared end cell equals cell open parentheses 11 over 2 close parentheses squared minus 4 open parentheses 5 over 2 close parentheses end cell row blank equals cell 121 over 4 minus 10 end cell row blank equals cell 81 over 4 end cell end table

Take the square root, remembering the plus-or-minus

alpha minus beta equals plus-or-minus 9 over 2

Finally, use the fact that alpha greater than beta

But  alpha greater than beta,  therefore  alpha minus beta greater than 0

bold italic alpha bold minus bold italic beta bold equals bold 9 over bold 2

(b) form a quadratic equation, with integer coefficients, which has roots  fraction numerator alpha plus beta over denominator alpha end fraction and fraction numerator alpha minus beta over denominator beta end fraction

Start by finding the sum and product of the new roots

table row Sum equals cell fraction numerator alpha plus beta over denominator a end fraction plus fraction numerator alpha minus beta over denominator beta end fraction end cell row blank equals cell fraction numerator beta open parentheses alpha plus beta close parentheses plus alpha open parentheses alpha minus beta close parentheses over denominator alpha beta end fraction end cell row blank equals cell fraction numerator alpha squared plus beta squared over denominator alpha beta end fraction end cell row blank equals cell fraction numerator open parentheses alpha squared plus 2 alpha beta plus beta squared close parentheses minus 2 alpha beta over denominator alpha beta end fraction end cell row blank equals cell fraction numerator open parentheses alpha plus beta close parentheses squared minus 2 alpha beta over denominator alpha beta end fraction end cell row blank equals cell fraction numerator open parentheses 11 over 2 close parentheses squared minus 2 open parentheses 5 over 2 close parentheses over denominator 5 over 2 end fraction end cell row blank equals cell 101 over 10 end cell row blank blank blank row blank blank blank row Product equals cell fraction numerator alpha plus beta over denominator a end fraction cross times fraction numerator alpha minus beta over denominator beta end fraction end cell row blank equals cell fraction numerator open parentheses alpha plus beta close parentheses open parentheses alpha minus beta close parentheses over denominator alpha beta end fraction end cell row blank equals cell fraction numerator open parentheses 11 over 2 close parentheses open parentheses 9 over 2 close parentheses over denominator 5 over 2 end fraction end cell row blank equals cell 99 over 10 end cell end table

Now use  Sum space of space roots equals negative b over a  and  Product space of space roots equals c over a

b over a equals negative 101 over 10 space space space and space space space c over a equals 99 over 10

Select  a equals 10  so b and c will have integer values

a equals 10 space space space space space space space space space b equals negative 101 space space space space space space space space space space c equals 99

And finally use those coefficients to write the new equation

bold 10 bold italic x to the power of bold 2 bold minus bold 101 bold italic x bold plus bold 99 bold equals bold 0

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Roger B

Author: Roger B

Expertise: Maths

Roger's teaching experience stretches all the way back to 1992, and in that time he has taught students at all levels between Year 7 and university undergraduate. Having conducted and published postgraduate research into the mathematical theory behind quantum computing, he is more than confident in dealing with mathematics at any level the exam boards might throw at you.

Dan Finlay

Author: Dan Finlay

Expertise: Maths Lead

Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.