Quadratic Formula and Nature of Roots

The Quadratic Formula

Applying completing-the-square to the general equation $ax^2 + bx + c = 0$ gives one master formula that solves every quadratic:

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

Here $a$, $b$, $c$ are read directly from the standard form. The $\pm$ gives the two roots.

Quick example

For $x^2 - 5x + 6 = 0$: $a = 1, b = -5, c = 6$. $$x = \frac{5 \pm \sqrt{25 - 24}}{2} = \frac{5 \pm 1}{2} = 3 \text{ or } 2.$$

Key Point: Always write the equation in standard form first, then substitute $a$, $b$, $c$ carefully (with signs) into the formula.

[Board Important] The formula works even when factorisation fails. It is the most reliable method — but watch your signs, especially for $b$.

The Discriminant

The quantity under the square root is the discriminant, denoted $D$ (or $\Delta$):

$$D = b^2 - 4ac$$

The discriminant alone — without solving — tells you the nature of the roots, because the $\sqrt{D}$ decides what kind of numbers the roots are.

Key Point: Compute $D = b^2 - 4ac$ first. Its sign instantly classifies the roots.

The three cases

Discriminant	Nature of roots
$D > 0$	Two distinct real roots
$D = 0$	Two equal real roots (one repeated)
$D < 0$	No real roots

[Board Important] 'Find the nature of the roots' means: compute $D$ and state which of the three cases applies. You do NOT need to find the roots themselves.

Using the Discriminant to Find Conditions

A very common exam type: 'Find the value of $k$ for which the equation has equal roots.' This uses $D = 0$.

Worked outline

For $x^2 - kx + 9 = 0$ to have equal roots: $D = 0$.

• $D = (-k)^2 - 4(1)(9) = k^2 - 36 = 0 \Rightarrow k^2 = 36 \Rightarrow k = \pm 6$.

Similarly:

• For real roots, require $D \geq 0$.
• For no real roots, require $D < 0$.

Key Point: Translate the condition on the roots into a condition on $D$, then solve for the unknown.

[Board Important] 'Equal roots' ⇒ $D = 0$; 'real and distinct' ⇒ $D > 0$; 'real roots' (allowing equal) ⇒ $D \geq 0$. Read the wording carefully.

Roots When They Are Equal

When $D = 0$, the two roots coincide and are both equal to:

$$x = -\frac{b}{2a}$$

(This is the formula with the $\sqrt{D} = 0$ term dropped.)

Example

$4x^2 - 12x + 9 = 0$: $D = 144 - 144 = 0$, so the roots are equal: $x = -\dfrac{-12}{2 \times 4} = \dfrac{12}{8} = \dfrac{3}{2}$.

Key Point: For equal roots, you don't need the full formula — just $x = -\dfrac{b}{2a}$.

[Board Important] When asked to find the equal root after showing $D = 0$, use $x = -\dfrac{b}{2a}$ directly. It saves time and avoids errors.

Solved Examples

Example 1: Apply the quadratic formula

Solve $2x^2 + x - 6 = 0$ using the formula.

Solution:

1. $a = 2, b = 1, c = -6$. $D = 1 - 4(2)(-6) = 1 + 48 = 49$.
2. $x = \dfrac{-1 \pm \sqrt{49}}{4} = \dfrac{-1 \pm 7}{4}$.
3. $x = \dfrac{6}{4} = \dfrac{3}{2}$ or $x = \dfrac{-8}{4} = -2$.

Final Answer: $x = \dfrac{3}{2}, -2$.

Takeaway: Compute $D$, then apply the $\pm$ formula.

Example 2: Find the discriminant

Find the discriminant of $2x^2 - 4x + 3 = 0$ and state the nature of roots.

Solution:

1. $a = 2, b = -4, c = 3$.
2. $D = (-4)^2 - 4(2)(3) = 16 - 24 = -8$.
3. $D < 0$.

Final Answer: $D = -8 < 0$, so no real roots.

Takeaway: Negative discriminant ⇒ no real roots.

Example 3: Roots with surds

Solve $x^2 - 4x + 1 = 0$ using the formula.

Solution:

1. $a = 1, b = -4, c = 1$. $D = 16 - 4 = 12$.
2. $x = \dfrac{4 \pm \sqrt{12}}{2} = \dfrac{4 \pm 2\sqrt3}{2} = 2 \pm \sqrt3$.

Final Answer: $x = 2 + \sqrt3, 2 - \sqrt3$.

Takeaway: Simplify $\sqrt{12} = 2\sqrt3$ and reduce the fraction.

Example 4: Equal roots — find $k$

Find $k$ so that $x^2 - kx + 9 = 0$ has equal roots.

Solution:

1. Equal roots ⇒ $D = 0$: $(-k)^2 - 4(1)(9) = 0$.
2. $k^2 - 36 = 0 \Rightarrow k = \pm 6$.

Final Answer: $k = 6$ or $k = -6$.

Takeaway: Equal roots ⇒ set $D = 0$ and solve.

Example 5: Distinct real roots condition

For what values of $k$ does $x^2 + kx + 4 = 0$ have two distinct real roots?

Solution:

1. Distinct real roots ⇒ $D > 0$: $k^2 - 16 > 0$.
2. $k^2 > 16 \Rightarrow k > 4$ or $k < -4$.

Final Answer: $k > 4$ or $k < -4$.

Takeaway: $D > 0$ gives an inequality in the unknown.

Example 6: Equal root value

Show that $4x^2 - 12x + 9 = 0$ has equal roots and find the root.

Solution:

1. $D = (-12)^2 - 4(4)(9) = 144 - 144 = 0$. So roots are equal.
2. Equal root: $x = -\dfrac{b}{2a} = -\dfrac{-12}{8} = \dfrac{3}{2}$.

Final Answer: Equal roots, $x = \dfrac{3}{2}$.

Takeaway: $D = 0$ ⇒ use $x = -b/(2a)$ for the repeated root.

Example 7: Formula with $a \neq 1$

Solve $3x^2 - 5x + 2 = 0$ using the formula.

Solution:

1. $a = 3, b = -5, c = 2$. $D = 25 - 24 = 1$.
2. $x = \dfrac{5 \pm 1}{6}$.
3. $x = 1$ or $x = \dfrac{4}{6} = \dfrac{2}{3}$.

Final Answer: $x = 1, \dfrac{2}{3}$.

Takeaway: A perfect-square discriminant gives rational roots.

Example 8: No real roots — find $k$ range

For what values of $k$ does $2x^2 + 3x + k = 0$ have no real roots?

Solution:

1. No real roots ⇒ $D < 0$: $9 - 8k < 0$.
2. $9 < 8k \Rightarrow k > \dfrac{9}{8}$.

Final Answer: $k > \dfrac{9}{8}$.

Takeaway: $D < 0$ gives the no-real-roots range.

Example 9: Discriminant decides a real-life feasibility

Can a rectangle have perimeter 20 m and area 30 m²? (Set up and use the discriminant.)

Solution:

1. Let length and breadth have sum $10$ (half-perimeter) and product 30. So they are roots of $t^2 - 10t + 30 = 0$.
2. $D = 100 - 120 = -20 < 0$.
3. No real roots, so such a rectangle is impossible.

Final Answer: No — the discriminant is negative, so no such rectangle exists.

Takeaway: A negative discriminant shows a situation has no real solution.

Example 10: Solve and simplify

Solve $x^2 + x - 1 = 0$ using the formula.

Solution:

1. $a = 1, b = 1, c = -1$. $D = 1 + 4 = 5$.
2. $x = \dfrac{-1 \pm \sqrt5}{2}$.

Final Answer: $x = \dfrac{-1 + \sqrt5}{2}, \dfrac{-1 - \sqrt5}{2}$.

Takeaway: When $D$ is not a perfect square, leave the roots in surd form.