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