Quadratics: solution of quadratic equations, nature of roots, quadratic inequalities

Quadratics: Solving Equations, Nature of Roots & Inequalities

1. Standard Form and Basic Definitions

A quadratic equation in one variable is written in the standard form

where \(a,b,c\) are real numbers. The graph of \(y=ax^{2}+bx+c\) is a parabola that opens upwards if \(a>0\) and downwards if \(a<0\).

2. Completing the Square and the Vertex Form

Completing the square is the key technique for deriving the quadratic formula, locating the vertex and solving equations.

1. Start with the standard form \(ax^{2}+bx+c=0\).
2. If \(aeq1\), divide the whole equation by \(a\):
   \(x^{2}+\frac{b}{a}x+\frac{c}{a}=0\)
3. Isolate the constant term:
   \(x^{2}+\frac{b}{a}x=-\frac{c}{a}\)
4. Add the square of half the coefficient of \(x\) to both sides:
   \(\displaystyle\left(\frac{b}{2a}\right)^{2}\)
5. Obtain the perfect‑square form:
   \(\displaystyle a\Bigl(x+\frac{b}{2a}\Bigr)^{2}= \frac{b^{2}-4ac}{4a}\)
6. Rewrite the left‑hand side as a vertex form:
   \(\displaystyle y=a\bigl(x-h\bigr)^{2}+k\), where \(h=-\frac{b}{2a}\) and \(k=\frac{4ac-b^{2}}{4a}\)

   The point \((h,k)\) is the **vertex** of the parabola.

3. Quadratic Formula and the Discriminant

Taking the square‑root of the completed‑square form and solving for \(x\) gives the quadratic formula:

The expression under the root, \(\displaystyle\Delta=b^{2}-4ac\), is called the discriminant. Its sign determines the nature of the roots.

Example – Using the Discriminant

Solve \(3x^{2}-12x+11=0\) and state the nature of its roots.

1. Coefficients: \(a=3,\;b=-12,\;c=11\).
2. Discriminant:
   \(\Delta = (-12)^{2}-4(3)(11)=144-132=12>0\)
3. \(\Delta>0\) ⇒ two distinct real roots.
4. Quadratic formula:
   \(x=\dfrac{12\pm\sqrt{12}}{6}= \dfrac{12\pm2\sqrt3}{6}=2\pm\frac{\sqrt3}{3}\)

4. Methods for Solving Quadratic Equations

Choose the quickest method for the given problem.

• Factorisation – when the quadratic can be expressed as a product of linear factors.
• Completing the square – useful for deriving the formula or when factorisation is not obvious.
• Quadratic formula – a universal method that works for any coefficients.

4.1 Example – Factorisation

Solve \(x^{2}-5x+6=0\).

\((x-2)(x-3)=0\) ⟹ \(x=2\) or \(x=3\).

4.2 Example – Completing the Square

Solve \(2x^{2}+8x+5=0\) by completing the square.

1. Divide by the leading coefficient \(2\): \(x^{2}+4x+\dfrac{5}{2}=0\).
2. Move the constant term: \(x^{2}+4x=-\dfrac{5}{2}\).
3. Add \((4/2)^{2}=4\) to both sides: \((x+2)^{2}= \dfrac{3}{2}\).
4. Take square‑roots: \(x+2 = \pm\sqrt{\dfrac{3}{2}}\).
5. Hence \(x = -2 \pm \sqrt{\dfrac{3}{2}}\).

4.3 Example – Quadratic Formula

Solve \(-4x^{2}+7x-2=0\).

Coefficients: \(a=-4,\;b=7,\;c=-2\). Discriminant: \(\Delta = 7^{2}-4(-4)(-2)=49-32=17\).

Applying the formula: \[ x=\frac{-7\pm\sqrt{17}}{-8}= \frac{7\mp\sqrt{17}}{8}. \]

5. Quadratic Inequalities

To solve an inequality of the form \(ax^{2}+bx+c\;\mathsf{op}\;0\) (\(\mathsf{op}\) is \(<,\,>,\,\le,\,\ge\)):

1. Find the real roots of the associated equation \(ax^{2}+bx+c=0\). (If \(\Delta<0\) the quadratic never changes sign.)
2. Note the sign of the leading coefficient \(a\).
3. Draw a sign chart (or test a point in each interval) to see where the quadratic is positive or negative.
4. Include the roots when the inequality is non‑strict (\(\le\) or \(\ge\)).

5.1 Example – Positive Leading Coefficient

Solve \(x^{2}-5x+6\ge0\).

• Factorise: \((x-2)(x-3)\ge0\). Roots: \(x=2,\;3\).
• Since \(a=1>0\), the quadratic is positive **outside** the roots.

Solution: \((-\infty,2]\cup[3,\infty)\).

5.2 Example – Negative Leading Coefficient

Solve \(-2x^{2}+3x-1\le0\).

1. Multiply by \(-1\) (reverse the inequality): \(2x^{2}-3x+1\ge0\).
2. Factorise: \((2x-1)(x-1)\ge0\). Roots: \(x=\tfrac12,\;1\).
3. Sign chart for \((2x-1)(x-1)\):

   Interval	Test point	\((2x-1)\)	\((x-1)\)	Product
   \((-\infty,\tfrac12)\)	0	–	–	+
   \([\tfrac12,1]\)	0.75	+	–	–
   \((1,\infty)\)	2	+	+	+

4. Because we reversed the inequality in step 1, the original inequality \(-2x^{2}+3x-1\le0\) is satisfied where the product is **negative or zero**: \([\tfrac12,1]\).

5.3 General Rules (Recap)

• If \(a>0\): the quadratic is positive outside the real roots and negative between them.
• If \(a<0\): the quadratic is negative outside the real roots and positive between them.
• If \(\Delta<0\): the sign of the quadratic is the sign of \(a\) for every real \(x\).

6. Simultaneous Equations Involving a Quadratic (Linear + Quadratic)

6.1 Method Overview

1. Solve the linear equation for one variable.
2. Substitute this expression into the quadratic equation – you obtain a single‑variable quadratic.
3. Solve the resulting quadratic (factorise, complete the square, or use the formula).
4. Back‑substitute each solution into the linear expression to get the corresponding value of the second variable.

6.2 Example 1 (Simple Substitution)

Solve

1. From the linear equation: \(y = 5 - 2x\).
2. Substitute into the quadratic: \[ x^{2}-(5-2x)=1 \;\Longrightarrow\; x^{2}+2x-6=0. \]
3. Factorise: \((x+3)(x-2)=0\) ⟹ \(x=-3\) or \(x=2\).
4. Find \(y\):
   • \(x=-3\): \(y=5-2(-3)=11\).
   • \(x=2\): \(y=5-2(2)=1\).

Solution set: \(\{(-3,\,11),\;(2,\,1)\}\).

6.3 Example 2 (Re‑arranging the Linear Equation)

Solve

1. Re‑arrange the linear equation for \(y\): \[ -2y = 7-3x \;\Longrightarrow\; y = \frac{3x-7}{2}. \]
2. Substitute into the quadratic: \[ x^{2}+\frac{3x-7}{2}=4 \;\Longrightarrow\; 2x^{2}+3x-7=8 \;\Longrightarrow\; 2x^{2}+3x-15=0. \]
3. Use the quadratic formula: \[ x=\frac{-3\pm\sqrt{3^{2}-4(2)(-15)}}{2\cdot2} =\frac{-3\pm\sqrt{129}}{4}. \]
4. Corresponding \(y\) values: \[ y=\frac{3x-7}{2} =\frac{3}{2}\!\left(\frac{-3\pm\sqrt{129}}{4}\right)-\frac{7}{2} =\frac{-9\pm3\sqrt{129}-14}{8} =\frac{-23\pm3\sqrt{129}}{8}. \]

Solution set: \[ \left\{\left(\frac{-3+\sqrt{129}}{4},\;\frac{-23+3\sqrt{129}}{8}\right),\; \left(\frac{-3-\sqrt{129}}{4},\;\frac{-23-3\sqrt{129}}{8}\right)\right\}. \]

7. Graphical Interpretation of Quadratics (Useful for the Syllabus)

• The discriminant \(\Delta\) tells how many \(x\)-intercepts the parabola has (0, 1 or 2).
• The sign of \(a\) tells whether the parabola opens **upwards** (\(a>0\)) or **downwards** (\(a<0\)).
• The vertex \((h,k)\) obtained from completing the square is the highest or lowest point of the graph, depending on the sign of \(a\).

Suggested diagram: a single sketch showing three parabolas side‑by‑side – one with \(\Delta>0\) (two intersections), one tangent (\(\Delta=0\)), and one with \(\Delta<0\) (no intersection). The opening direction is indicated by the sign of \(a\).

8. Common Pitfalls & Examination Tips

• Factor first. A quick factorisation saves time and avoids arithmetic errors.
• Keep the “divide by \(a\)” step visible when completing the square – it prevents sign slips.
• Check the discriminant early. Its sign tells you instantly whether real roots exist and which method is most efficient.
• In inequalities:
  • Remember to include the roots for “\(\le\)” or “\(\ge\)”.
  • If \(a<0\), the sign pattern is the reverse of the case \(a>0\).
• For simultaneous equations, write the linear expression clearly before substitution; a messy algebraic step is a common source of loss of marks.
• If the roots are complex, express them as \(p\pm qi\) where \(q=\dfrac{\sqrt{-\Delta}}{2a}\).
• When using the quadratic formula, always simplify the fraction fully – unnecessary radicals cost marks.

9. Quick Reference Summary