Roots of Quadratic Equations (Edexcel IGCSE Further Pure Maths): Revision Note

Discriminants

What is the discriminant of a quadratic function?

• The discriminant of a quadratic is often denoted by the Greek letter $\mathrm{\Delta }$ (upper case delta)

• For a quadratic  $a{x}^2+bx+c\left(a\ne 0\right)$ the discriminant is given by

  • $\mathrm{\Delta }=b^2-4ac$ 

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

  • $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

• 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  $a{x}^2+bx+c=0$

  • It also tells us about the graph of  $y=a{x}^2+bx+c$

• If Δ > 0 then $\sqrt{b^2-4ac}$ and $-\sqrt{b^2-4ac}$ are two distinct values

  • The equation $a{x}^2+bx+c=0$ has unequal real roots

    • i.e. there are two distinct real solutions

  • The graph of $y=a{x}^2+bx+c$ crosses the x-axis twice

• If $\mathrm{\Delta }=0$ then $\sqrt{b^2-4ac}$ and $-\sqrt{b^2-4ac}$ are both zero

  • The equation $a{x}^2+bx+c=0$ has equal real roots

    • i.e. it has one repeated real solution

  • The graph of $y=a{x}^2+bx+c$ touches the x-axis at exactly one point

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

• If $\mathrm{\Delta }<0$ then $\sqrt{b^2-4ac}$ and $-\sqrt{b^2-4ac}$ are both undefined

  •   The roots of the equation $a{x}^2+bx+c=0$ are not real

    • i.e. it has no real solutions

  • The graph of $y=a{x}^2+bx+c$ never touches the x-axis

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

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 $\mathrm{\Delta }=b^2-4ac$

  • Decide whether $\mathrm{\Delta }>0$, $\mathrm{\Delta }=0$ or $\mathrm{\Delta }<0$

    • If the question says there are real roots but does not specify how many then use $\mathrm{\Delta }\ge 0$

  • Solve the resulting equation or inequality for $k$

Sum & Product of Roots

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

• A quadratic equation $a{x}^2+bx+c=0$ (where $a\ne 0$) has roots $\alpha$ and $\beta$ given by

  • $\alpha ,\beta =\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

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

• This means the equation can be rewritten in the form  $a(x-\alpha )(x-\beta )=0$

  • Note that $(x-\alpha )(x-\beta )=x^2-(\alpha +\beta )x+\alpha \beta$

    • It is possible that the roots are repeated, i.e. that $\alpha =\beta$

  • You can then equate the two forms:

    • $a{x}^2+bx+c=a(x-\alpha )(x-\beta )$

  • Then (because $a\ne 0$) you can divide both sides of that by a and expand the brackets:

    • $x^2+\frac{b}{a}x+\frac{c}{a}=x^2-(\alpha +\beta )x+\alpha \beta$

  • Finally, compare the coefficients

    • $x$ coefficients:  $-\frac{b}{a}=(\alpha +\beta )$

    • Constant terms:  $\frac{c}{a}=\alpha \beta$

• Therefore for a quadratic equation $a{x}^2+bx+c=0$  :

  • The sum of the roots $\alpha +\beta$ is equal to $-\frac{b}{a}$

  • The product of the roots $\alpha \beta$ is equal to $\frac{c}{a}$

• 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 $a$, $b$ 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  $\alpha +\beta =\frac{7}{2}$  and  $\alpha \beta =-2$. Find a quadratic equation with integer coefficients that has roots $\alpha$ and $\beta$."

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

  • $\alpha +\beta =-\frac{b}{a}$ 

    • So  $\frac{b}{a}=-\frac{7}{2}$

  • $\alpha \beta =\frac{c}{a}$

    • So  $\frac{c}{a}=-2$

• 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 $\alpha +\beta$ and $\alpha \beta$ into integers

    • Let $a=2$

    • Then  $b=-\frac{7}{2}\times 2=-7$

    • And  $c=-2\times 2=-4$

• STEP 3
  Write down the equation using your values for $a$, $b$ and $c$

  • $2x^2-7x-4=0$

• Note that the answer is not unique

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

    • e.g.  $4x^2-14x-8=0$

How do I find a quadratic equation with related roots?

• 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 $\alpha +\beta$ and $\alpha \beta$

  • $\alpha$ and $\beta$ are the roots of the first quadratic

  • For example:
    

    • $\begin{array}{rcl}{\alpha }^2+{\beta }^2& =& {\alpha }^2+2\alpha \beta +{\beta }^2-2\alpha \beta \\ & =& {\left(\alpha +\beta \right)}^2-2\alpha \beta \end{array}$
      

    • $\begin{array}{rcl}{\alpha }^3+{\beta }^3& =& {\alpha }^3+3\alpha \beta \left(\alpha +\beta \right)+{\beta }^3-3\alpha \beta \left(\alpha +\beta \right)\\ & =& {\alpha }^3+3{\alpha }^2\beta +3\alpha {\beta }^2+{\beta }^3-3\alpha \beta \left(\alpha +\beta \right)\\ & =& {\left(\alpha +\beta \right)}^3-3\alpha \beta \left(\alpha +\beta \right)\end{array}$
      

    • $\begin{array}{rcl}{\left(\alpha -\beta \right)}^2& =& {\alpha }^2-2\alpha \beta +{\beta }^2\\ & =& {\alpha }^2+2\alpha \beta +{\beta }^2-4\alpha \beta \\ & =& {\left(\alpha +\beta \right)}^2-4\alpha \beta \end{array}$
      

• Then if you know the values of $\alpha +\beta$ 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  $\alpha +\beta =-\frac{3}{2}$  and  $\alpha \beta =2$.  Form a second quadratic equation with integer coefficients that has roots $\frac{\alpha }{{\beta }^2}$ and $\frac{\beta }{{\alpha }^2}$."

  • The sum of the new roots is  $\frac{\alpha }{{\beta }^2}+\frac{\beta }{{\alpha }^2}=\frac{{\alpha }^3+{\beta }^3}{{\alpha }^2{\beta }^2}=\frac{{\alpha }^3+{\beta }^3}{{\left(\alpha \beta \right)}^2}$

    • We can use the substitution for ${\alpha }^3+{\beta }^3$ from above

    • So  $\mathrm{Sum}=\frac{{\left(\alpha +\beta \right)}^3-3\alpha \beta \left(\alpha +\beta \right)}{{\left(\alpha \beta \right)}^2}=\frac{{\left(-{\displaystyle \frac{3}{2}}\right)}^3-3\left(2\right)\left(-{\displaystyle \frac{3}{2}}\right)}{{\left(2\right)}^2}=\frac{45}{32}$

  • The product of the new roots is  $\frac{\alpha }{{\beta }^2}\times \frac{\beta }{{\alpha }^2}=\frac{\alpha \beta }{{\alpha }^2{\beta }^2}=\frac{1}{\alpha \beta }$

    • So  $\mathrm{Product}=\frac{1}{\alpha \beta }=\frac{1}{2}$

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

    • $32x^2-45x+16=0$