Find the solution set for quadratic inequalities either graphically or algebraically, expressing solutions using correct inequality notation

Quadratic Functions and Inequalities (IGCSE Additional Mathematics 0606 – Section 2)

Learning Objectives

• Write a quadratic function in standard, factorised and vertex (completed‑square) forms.
• Find the maximum or minimum value and the corresponding x‑coordinate by completing the square or by differentiation.
• Determine the real roots of the associated quadratic equation using factorisation, the quadratic formula or completing the square.
• Analyse the sign of a quadratic expression and solve the inequalities

  ax²+bx+c < 0, ≤ 0, > 0, ≥ 0

  both graphically and algebraically.

• Express solution sets correctly using inequality notation and interval notation.
• Understand how discriminant conditions relate to the number of intersections between a parabola and a straight line.

1. Forms of a Quadratic Function

Derivation of the vertex formula (completing the square):

\[

\begin{aligned}

f(x)&=ax^{2}+bx+c

=a\Bigl[x^{2}+\frac{b}{a}x\Bigr]+c\\[2mm]

&=a\Bigl[\bigl(x+\tfrac{b}{2a}\bigr)^{2}-\frac{b^{2}}{4a^{2}}\Bigr]+c\\[2mm]

&=a\bigl(x+\tfrac{b}{2a}\bigr)^{2}+ \Bigl(c-\frac{b^{2}}{4a}\Bigr).

\end{aligned}

\]

Hence the vertex is \((h,k)=\bigl(-\tfrac{b}{2a},\,c-\tfrac{b^{2}}{4a}\bigr)\).

Maximum / Minimum (Vertex) – Differentiation Alternative

• Differentiate: \(f'(x)=2ax+b\).
• Set \(f'(x)=0\) ⇒ \(x=-\dfrac{b}{2a}\) (the same abscissa as above).
• Substitute this \(x\) into \(f(x)\) to obtain the ordinate \(k\).
• Use the sign of \(a\) to decide whether the vertex is a minimum (\(a>0\)) or a maximum (\(a<0\)).

2. The Discriminant and Roots

When a straight line \(y=mx+c\) is involved, replace the line by \(0\) and use the same discriminant test on the resulting quadratic equation to decide whether the line cuts the parabola twice, touches it, or misses it.

3. Sign Pattern of a Quadratic

Because a quadratic can change sign only at its real roots, the sign on each interval is completely determined by the leading coefficient \(a\).

• If \(a>0\) (parabola opens upward): + – +

  (positive outside the roots, negative between them).

• If \(a<0\) (parabola opens downward): – + –

  (negative outside, positive between).

Why this works: The quadratic is continuous. Starting far to the left, its sign equals the sign of \(a\) (the dominant term \(ax^{2}\)). Each time we cross a simple real root the sign flips; a double root does not change the sign.

4. Solving Quadratic Inequalities

4.1 Graphical Method

1. Rewrite the inequality as \(f(x)\;\#\;0\) where \(\#\) is <, ≤, > or ≥.
2. Sketch the graph of \(y=f(x)\):

   • Find the vertex \((h,k)\) and the direction of opening.
   • Locate the x‑intercepts (real roots) – they split the x‑axis into intervals.
   • Mark the axis of symmetry \(x=h\) for accuracy.

3. Shade the region that satisfies the inequality:

   • Above the x‑axis → \(f(x)>0\).
   • Below the x‑axis → \(f(x)<0\).
   • On the axis → include the points only for “≤” or “≥”.

4. Translate the shaded part(s) into inequality form and interval notation.

4.2 Algebraic Method (Sign‑Chart)

1. Standardise the inequality so that all terms are on the left: \(ax^{2}+bx+c\;\#\;0\).
2. Compute the discriminant \(D=b^{2}-4ac\) and obtain the real roots (if any) using the quadratic formula or factorisation.
3. Construct a sign chart:

   Interval	Test point	Sign of \(f(x)\)
   \((-\infty,r{1})\)	any \(x{1}\)	sign of \(a\)
   \((r{1},r{2})\)	any \(r{1}{2}\)	opposite to sign of \(a\)
   \((r{2},\infty)\)	any \(x>r{2}\)	sign of \(a\)

   If \(D=0\) there are only two intervals; if \(D<0\) there is a single interval \((-\infty,\infty)\) whose sign equals that of \(a\).

4. Choose the intervals that satisfy the required inequality:

   • For “<” or “>” exclude the root points.
   • For “≤” or “≥” include the root points.

5. Write the solution set:

   • In inequality form (e.g. \(x{1}\) or \(x>r{2}\)).
   • In interval notation (e.g. \((-\infty,r{1})\cup(r{2},\infty)\)).

5. Graph‑Sketch Checklist (Useful for Both Methods)

• Identify the leading coefficient \(a\) → direction of opening.
• Find the vertex \((h,k)\) using either completing the square or differentiation.
• Calculate the discriminant \(D\) → number of real roots.
• Obtain the x‑intercepts (if \(D\ge0\)).
• Mark the axis of symmetry \(x=h\).
• State the domain (\(\mathbb{R}\)) and the range:

  • If \(a>0\): range \([k,\infty)\).
  • If \(a<0\): range \((-\infty,k]\).

6. Worked Examples

Example 1 – Graphical Solution of a Strict Inequality

Solve \(x^{2}-4x-5<0\).

1. Write \(f(x)=x^{2}-4x-5\).
2. Factorise: \((x-5)(x+1)=0\) → roots \(r{1}=-1,\;r{2}=5\).
3. Since \(a=1>0\) the parabola opens upward; sign pattern is + – +.
4. Below the x‑axis between the roots → solution \(-1

5. On a graph, shade the portion of the curve that lies under the axis and note that the axis itself is not included.

Example 2 – Algebraic Solution of a Non‑Strict Inequality

Solve \(2x^{2}+3x-2\ge0\).

1. Standard form already: \(f(x)=2x^{2}+3x-2\).
2. Discriminant: \(D=3^{2}-4\cdot2\cdot(-2)=9+16=25>0\).
3. Roots: \(x=\dfrac{-3\pm5}{4}\) ⇒ \(r{1}=-2,\;r{2}= \dfrac12\).
4. Sign pattern for \(a>0\) is + – +.
5. “≥0” requires the outer intervals including the roots:

Solution: \(x\le-2\) or \(x\ge\frac12\) or \((-\infty,-2]\cup[\tfrac12,\infty)\).

Example 3 – No Real Roots

Solve \(-3x^{2}+12x-9\le0\).

1. Factor: \(-3(x^{2}-4x+3)=-3(x-1)(x-3)\).
2. Discriminant \(D=12^{2}-4(-3)(-9)=144-108=36>0\) → roots \(1\) and \(3\).
3. Because \(a=-3<0\), sign pattern is – + –.
4. For “≤0” we take the outer intervals including the roots:

Solution: \(x\le1\) or \(x\ge3\) or \((-\infty,1]\cup[3,\infty)\).

Example 4 – Using the Discriminant for a Line‑Parabola Intersection

Find the values of \(k\) for which the line \(y=k\) meets the parabola \(y=x^{2}+4x+5\) in:

• two points,
• exactly one point,
• no point.

Set \(x^{2}+4x+5=k\) ⇒ \(x^{2}+4x+(5-k)=0\).

Discriminant \(D=4^{2}-4\cdot1\cdot(5-k)=16-20+4k=4k-4\).

• \(D>0\) ⇒ \(4k-4>0\) ⇒ \(k>1\) → two intersections.
• \(D=0\) ⇒ \(k=1\) → the line is tangent.
• \(D<0\) ⇒ \(k<1\) → no intersection.

Example 5 – Quadratic Formula When Factorisation Is Not Easy

Solve \(3x^{2}+2x-7=0\) using the quadratic formula.

\[

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

=\frac{-2\pm\sqrt{2^{2}-4\cdot3\cdot(-7)}}{2\cdot3}

=\frac{-2\pm\sqrt{4+84}}{6}

=\frac{-2\pm\sqrt{88}}{6}

=\frac{-2\pm2\sqrt{22}}{6}

=\frac{-1\pm\sqrt{22}}{3}.

\]

7. Common Mistakes & How to Avoid Them

• Omitting boundary points for “≤” or “≥”. Always substitute the root back into the original inequality to check.
• Reversing the sign pattern when \(a<0\). Remember the pattern is “– + –”.
• Mis‑interpreting the discriminant: \(D<0\) means *no* real roots, so the quadratic never changes sign – it keeps the sign of \(a\) for all real \(x\).
• Choosing a test point that coincides with a root. Test points must lie strictly inside an interval.
• Graphical errors: forget to draw the axis of symmetry or to shade the axis itself for non‑strict inequalities.

8. Practice Questions

1. Solve \(x^{2}+6x+5>0\). Write the answer in both inequality and interval notation.
2. Find the solution set of \(-3x^{2}+12x-9\le0\).
3. Determine all \(x\) for which \(4x^{2}-4x-15<0\).
4. Using a sign chart, explain why \(x^{2}+1\ge0\) holds for every real \(x\).
5. For the quadratic \(f(x)=2x^{2}-8x+6\):

   • Write it in vertex form.
   • State the minimum value and the x‑coordinate at which it occurs.
   • Solve the inequality \(f(x)\le4\).

9. Connections to Other Sections

• Section 3 (Simultaneous Equations): Solutions of quadratic equations are often required when solving simultaneous linear‑quadratic systems.
• Section 14 (Calculus): The vertex obtained by differentiation is a stationary point; later you will use the first derivative test to classify maxima and minima.
• Section 5 (Coordinate Geometry): The discriminant determines the number of intersection points between a line and a parabola.

10. Summary Checklist

• Convert the inequality to the form \(ax^{2}+bx+c\;\#\;0\).
• Compute the discriminant \(D\) and obtain real roots (if any) using factorisation or the quadratic formula.
• Identify the sign pattern from the leading coefficient \(a\) (or verify with a test point).
• Select the intervals that satisfy the required inequality; remember to include the roots for “≤” or “≥”.
• Express the final answer clearly in inequality form and/or interval notation.
• If a maximum/minimum is required, give the vertex \((h,k)\) and indicate whether it is a maximum (\(a<0\)) or a minimum (\(a>0\)).