Roots of Quadratic Equations (Edexcel IGCSE Further Maths)

Discriminants

Spec point — 2B

What is the discriminant of a quadratic function?

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

• For a quadratic $a x^{2} + b x + c ( a \text{eq} 0 )$ the discriminant is given by

  • $Δ=b^{2}-4ac$

• 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 $ax^{2}+bx+c=0$

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

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

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

    • i.e. there are two distinct real solutions

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

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

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

    • i.e. it has one repeated real solution

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

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

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

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

    • i.e. it has no real solutions

  • The graph of $y=ax^{2}+bx+c$ 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 $Δ=b^{2}-4ac$

  • Decide whether $Δ>0$, $Δ=0$ or $Δ<0$

    • If the question says there are real roots but does not specify how many then use $Δ\geq0$

  • Solve the resulting equation or inequality for $k$

Exam Hint

• 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 $f(x)=2kx^{2}+kx-k+2$ , where $k$ is a constant. The graph of $y=f(x)$ intercepts the $x$-axis at two different points.

a) Show that $9k^{2}-16k>0$.

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

This means that the discriminant $b^{2}-4ac$ is greater than zero

Here $a=2k$, $b=k$, and $c=-k+2$

$( k )^{2} - 4 ( 2 k ) ( - k + 2 ) > 0$

Expand the brackets and collect terms

$\begin{gathered} k^{2} - 8 k ( - k + 2 ) & > 0 \\ k^{2} + 8 k^{2} - 16 k & > 0 \end{gathered}$

$9k^{2}-16k>0$

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

Solve the inequality, beginning by factorising

$k ( 9 k - 16 ) > 0$

This tells us the graph of $y=9k^{2}-16k$ intercepts the horizontal axis at $k=0$ and $k=\frac{16}{9}$

It can be helpful to sketch a graph here

$9k^{2}-16k>0$ will be true to the left of 0 and to the right of $\frac{16}{9}$

Write these down as inequalities

$k<0\mathrm{or}k>\frac{16}{9}$

Sum & Product of Roots

Spec point — 2C

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

• A quadratic equation $ax^{2}+bx+c=0$ (where $a\neq0$) has roots $α$ and $β$ given by

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

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

• This means the equation can be rewritten in the form $a(x-α)(x-β)=0$

  • Note that $(x-α)(x-β)=x^{2}-(α+β)x+αβ$

    • It is possible that the roots are repeated, i.e. that $α=β$

  • You can then equate the two forms:

    • $ax^{2}+bx+c=a(x-α)(x-β)$

  • Then (because $a\neq0$) you can divide both sides of that by a and expand the brackets:

    • $x^{2}+\frac{b}{a}x+\frac{c}{a}=x^{2}-(α+β)x+αβ$

  • Finally, compare the coefficients

    • $x$ coefficients: $-\frac{b}{a}=(α+β)$
    • Constant terms: $\frac{c}{a}=αβ$

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

  • The sum of the roots $α+β$ is equal to $-\frac{b}{a}$
  • The product of the roots $αβ$ 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 $α$ and $β$, where $α+β=\frac{7}{2}$ and $αβ=-2$. Find a quadratic equation with integer coefficients that has roots $α$ and $β$."

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

  • $α+β=-\frac{b}{a}$

    • So $\frac{b}{a}=-\frac{7}{2}$

  • $αβ=\frac{c}{a}$

    • 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 $α+β$ and $αβ$ into integers

    • Let $a=2$
    • Then $b=-\frac{7}{2}\times2=-7$
    • And $c=-2\times2=-4$

• 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 $α$ and $β$ in terms of $α+β$ and $αβ$

  • $α$ and $β$ are the roots of the first quadratic

  • For example:

    • $\begin{gathered} \\alpha ^{2} + \\beta ^{2} & = \\alpha ^{2} + 2 \\alpha \\beta + \\beta ^{2} - 2 \\alpha \\beta \\ = ( \\alpha + \\beta )^{2} - 2 \\alpha \\beta \end{gathered}$
    • $\begin{gathered} \\alpha ^{3} + \\beta ^{3} & = \\alpha ^{3} + 3 \\alpha \\beta ( \\alpha + \\beta ) + \\beta ^{3} - 3 \\alpha \\beta ( \\alpha + \\beta ) \\ = \\alpha ^{3} + 3 \\alpha ^{2} \\beta + 3 \\alpha \\beta ^{2} + \\beta ^{3} - 3 \\alpha \\beta ( \\alpha + \\beta ) \\ = ( \\alpha + \\beta )^{3} - 3 \\alpha \\beta ( \\alpha + \\beta ) \end{gathered}$
    • $\begin{gathered} ( \\alpha - \\beta )^{2} & = \\alpha ^{2} - 2 \\alpha \\beta + \\beta ^{2} \\ = \\alpha ^{2} + 2 \\alpha \\beta + \\beta ^{2} - 4 \\alpha \\beta \\ = ( \\alpha + \\beta )^{2} - 4 \\alpha \\beta \end{gathered}$

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

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

    • We can use the substitution for $α^{3}+β^{3}$ from above
    • So $Sum = \frac{ ( \alpha + \beta ) ^{3} - 3 \alpha \beta ( \alpha + \beta ) }{ ( \alpha \beta ) ^{2} } = \frac{ ( - \text{begin} \text{display} \text{style} \frac{3}{2} ) ^{3} - 3 ( 2 ) ( - \text{begin} \text{display} \text{style} \frac{3}{2} ) }{ ( 2 ) ^{2} } = \frac{45}{32}$

  • The product of the new roots is $\frac{α}{β^{2}}\times\frac{β}{α^{2}}=\frac{αβ}{α^{2}β^{2}}=\frac{1}{αβ}$

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

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

    • $32x^{2}-45x+16=0$

Worked Example The roots of the quadratic equation $2x^{2}-11x+5=0$ are $α$ and $β$.

Given that $α>β$ and without solving the equation,

(a) show that $α-β=\frac{9}{2}$

Use $α+β=-\frac{b}{a}$ and $αβ=\frac{c}{a}$ to find the sum and product

$\\alpha + \\beta = - \frac{( - 11 )}{2} = \frac{11}{2}$

$αβ=\frac{5}{2}$

Express $( \\alpha - \\beta )^{2}$ in terms of $α+β$ and $αβ$

$\begin{gathered} ( \\alpha - \\beta )^{2} & = \\alpha ^{2} - 2 \\alpha \\beta + \\beta ^{2} \\ = ( \\alpha ^{2} + 2 \\alpha \\beta + \\beta ^{2} ) - 4 \\alpha \\beta \\ = ( \\alpha + \\beta )^{2} - 4 \\alpha \\beta \end{gathered}$

Substitute to find the value

$\begin{gathered} ( \\alpha - \\beta )^{2} & = ( \frac{11}{2} )^{2} - 4 ( \frac{5}{2} ) \\ = \frac{121}{4} - 10 \\ = \frac{81}{4} \end{gathered}$

Take the square root, remembering the $\pm$

$α-β=\pm\frac{9}{2}$

Finally, use the fact that $α>β$

But $α>β$, therefore $α-β>0$

$α-β=\frac{9}{2}$

(b) form a quadratic equation, with integer coefficients, which has roots $\frac{α+β}{α}$ and $\frac{α-β}{β}$

Start by finding the sum and product of the new roots

| Sum=(α+β)/(a)+(α - β)/(β) | | _=(β (α+β)+α (α - β))/(α β) | | _=(α ^2+β ^2)/(α β) | | _=((α ^2+2α β+β ^2) - 2α β)/(α β) | | _=((α+β) ^2 - 2α β)/(α β) | | _=((11 / 2) ^2 - 2 (5 / 2))/(5 / 2) | | _=101 / 10 | | _ _ _ | _ _ _ | Product=(α+β)/(a) x (α - β)/(β) | | _=((α+β)(α - β))/(α β) | | _=((11 / 2)(9 / 2))/(5 / 2) | | _=99 / 10 | |

Now use $\mathrm{Sum} \mathrm{of} \mathrm{roots}=-\frac{b}{a}$ and $\mathrm{Product} \mathrm{of} \mathrm{roots}=\frac{c}{a}$

$\frac{b}{a}=-\frac{101}{10}\mathrm{and}\frac{c}{a}=\frac{99}{10}$

Select $a=10$ so $b$ and $c$ will have integer values

$a=10b=-101c=99$

And finally use those coefficients to write the new equation

$10x^{2}-101x+99=0$