Quadratic Equations

🔹 Definition

A quadratic equation in one variable is an equation of the form:

$$ax^2 + bx + c = 0$$

where

• x is a variable

• $a, b, c$ are real numbers, and

• $a \ne 0$.

🔹 Standard Form

$$ax^2 + bx + c = 0$$

Here:

• Coefficient of $x^2 = a$
  

• Coefficient of $x = b$
  

• Constant term $= c$
  

🔹 Methods of Solving a Quadratic Equation

🧭 Method 1: Factorisation (Splitting the Middle Term)

✅ Stepwise Procedure

Step 1: Write the equation in standard form:

$$ax^2 + bx + c = 0$$

Step 2: Identify coefficients.

• x=variable

• Coefficient of $x^2 = a$
  

• Coefficient of $x = b$
  

• Constant term $= c$
  

Step 3: Multiply

$$(\text{Coefficient of } x^2) \times (\text{Constant term}) = a \times c$$

Step 4: Find two numbers $m$ and $n$ such that

$$m + n = b, \quad m \times n = a \times c$$

Step 5: Rewrite the middle term $bx$ as $mx + nx$.

Step 6: Group and factorise the terms.

Step 7: Equate each factor to zero and find the value of $x$.

🧩 Rule Box: Splitting the Middle Term

Find two numbers $m$ and $n$ such that

$$m + n = \text{Coefficient of } x$$

$$$$and also
$$m \times n = (\text{Coefficient of } x^2) \times (\text{Constant term})$$Then rewrite $bx$ as $mx + nx$,

Now group and factorise.

✳️ Example 1 (Easy)

Solve $x^2 + 5x + 6 = 0$

Step 1: Compare the Coefficients of the given qudratic equation with the standard quadratic equation

$$a{x}^2+bx+c=0$$$\mathrm{On\; comparision\; we\; get:}$

$a = 1, b = 5, c = 6$

$$Step 2: Multiply $a \times c = 1 \times 6 = 6$

Step 3: Find $m, n$ such that:
$m + n = 5, \quad m \times n = 6$
So, $m = 2, n = 3$

Step 4: Rewrite middle term:

$$x^2 + 2x + 3x + 6 = 0$$

Step 5: Group and factorise:

$$x(x + 2) + 3(x + 2) = 0$$
$$(x + 2)(x + 3) = 0$$Step 6:

$$x + 2 = 0 \Rightarrow x = -2 \quad \text{and} \quad x + 3 = 0 \Rightarrow x = -3$$

✅ Roots: $x = -2, -3$

✳️ Example 2 (Moderate)

Solve $2x^2 + 5x + 2 = 0$

Step 1:

Compare the Coefficients of the given qudratic equation with the standard quadratic equation

$$a{x}^2+bx+c=0$$$\mathrm{On\; comparision\; we\; get:}$

$a = 2, b = 5, c = 2$

Step 2: $a \times c = 4$

Step 3: Find $m, n$ such that:
$m+n=5,m\times n=4$

$m + n = 5, m \times n = 4 \Rightarrow m = 4, n = 1$

Step 4: Rewrite middle term:

$$2x^2 + 4x + x + 2 = 0$$

Step 5: Group and factorise:

$$2x(x + 2) + 1(x + 2) = 0$$
$$(x + 2)(2x + 1) = 0$$Step 6:

$$x = -2 \quad \text{or} \quad x = -\frac{1}{2}$$

✅ Roots: $x = -2, -\frac{1}{2}$

✳️ Example 3 (Hard)

Solve $3x^2 – 8x – 3 = 0$

Step 1:

Compare the Coefficients of the given qudratic equation with the standard quadratic equation

$$a{x}^2+bx+c=0$$$\mathrm{On\; comparision\; we\; get:}$

$a = 3, b = -8, c = -3$

Step 2: $a \times c = -9$

Step 3: Find $m, n$ such that:
$m + n = -8, m \times n = -9$
$m = -9, n = 1$

Step 4: Rewrite middle term:

$$3x^2 – 9x + x – 3 = 0$$

Step 5: Group and factorise:

$$3x(x – 3) + 1(x – 3) = 0$$

$$$$$$(x – 3)(3x + 1) = 0$$

Step 6:

$$x = 3, \quad x = -\frac{1}{3}$$

✅ Roots: $x = 3, -\frac{1}{3}$

Method 2: Completing the Square

🧭 Concept

A quadratic equation

$$ax^2 + bx + c = 0$$

can be solved by expressing it as a perfect square of a binomial.

✅ Step-by-Step Procedure

Step 1: Bring the equation to the form

$$ax^2 + bx = -c$$

Step 2: Divide each term by $a$ to make the coefficient of $x^2$ equal to 1.

$$x^2+\frac{b}{a}x=-\frac{c}{a}$$Step 3: Add $\left(\frac{b}{2a}\right)^2$to both sides to make a perfect square.

Step 4: Write the LHS as a square:

$${(x+\frac{b}{2a})}^2=\frac{b^2-4ac}{4a^2}$$Step 5: Take square root on both sides.

Step 6: Solve for $x$.

🧩 Rule Box: Completing the Square

To complete the square for $x^2 + px$,
add and subtract $\left(\frac{p}{2}\right)^2$.

$$x^2 + px + \left(\frac{p}{2}\right)^2 = \left(x + \frac{p}{2}\right)^2$$

✳️ Example 1 (Easy)

Solve $x^2 + 6x + 5 = 0$

Step 1: Move constant to RHS:

$$x^2 + 6x = -5$$

Step 2: Add $\left(\frac{6}{2}\right)^2 = 9$ to both sides:

$$x^2+6x+9=-5+9$$

Step 3: Write as perfect square:

$$(x+3{)}^2=4$$

Step 4: Take square roots:

$$x + 3 = \pm 2$$

Step 5:

$$x = -3 \pm 2$$

✅ Roots: $x = -1, -5$

✳️ Example 2 (Moderate)

Solve $2x^2 + 3x – 2 = 0$

Step 1: Divide by 2:$$x^2 + \frac{3}{2}x – 1 = 0$$

Step 2: Move constant:

$$x^2+\frac{3}{2}x=1$$

Step 3: Add $\left(\frac{3}{4}\right)^2 = \frac{9}{16}$:

$$x^2 + \frac{3}{2}x + \frac{9}{16} = 1 + \frac{9}{16}$$

Step 4: Write as square:

$$\left(x + \frac{3}{4}\right)^2 = \frac{25}{16}$$

Step 5: Take square roots:

$$x + \frac{3}{4} = \pm \frac{5}{4}$$

Step 6:

$$x = \frac{-3 \pm 5}{4}$$

✅ Roots: $x = \tfrac{1}{2}, -2$.

✳️ Example 3 (Hard)

Solve $3x^2 – 10x + 7 = 0$

Step 1: Divide by 3:

$$x^2 – \frac{10}{3}x + \frac{7}{3} = 0$$

Step 2: Move constant:

$$x^2 – \frac{10}{3}x = -\frac{7}{3}$$

Step 3: Add $\left(\frac{-10}{6}\right)^2 = \frac{25}{9}$:

$$x^2 – \frac{10}{3}x + \frac{25}{9} = -\frac{7}{3} + \frac{25}{9}$$

Step 4: Simplify RHS:

$$-\frac{7}{3} + \frac{25}{9} = \frac{4}{9}$$$$\left(x – \frac{5}{3}\right)^2 = \frac{4}{9}$$

Step 5:

$$x – \frac{5}{3} = \pm \frac{2}{3}$$

Step 6:

$$x = \frac{5 \pm 2}{3}$$

✅ Roots: $x = \frac{7}{3}, 1$

🔹 Method 3: Quadratic Formula

🧭 Concept

For any quadratic equation

$$ax^2 + bx + c = 0,$$

the solution is given by the quadratic formula:

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

Here,

$$D = b^2 – 4ac$$

s called the Discriminant.

✅ Step-by-Step Procedure

Step 1: Identify $a, b, c$.
Step 2: Compute discriminant $D = b^2 – 4ac$.
Step 3: Substitute in the formula

$$x = \frac{-b \pm \sqrt{D}}{2a}$$

Step 4: Simplify to get the roots.

🧩 Rule Box: Quadratic Formula

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

✳️ Example 1 (Easy)

Solve $x^2 – 5x + 6 = 0$

Step 1: $a = 1, b = -5, c = 6$

Step 2: $D = (-5)^2 – 4(1)(6) = 25 – 24 = 1$

Step 3:

$$x = \frac{-(-5) \pm \sqrt{1}}{2(1)} = \frac{5 \pm 1}{2}$$

Step 4:

$$x = 3 \text{ or } 2$$

✅ Roots: $x = 2, 3$.

✳️ Example 2 (Moderate)

Solve $2x^2 + 3x – 2 = 0$

Step 1: $a = 2, b = 3, c = -2$

Step 2: $D = 3^2 – 4(2)(-2) = 9 + 16 = 25$

Step 3:

$$x = \frac{-3 \pm \sqrt{25}}{4}$$

Step 4:

$$x = \frac{-3 \pm 5}{4}$$

Step 5:

$$x = \frac{1}{2}, -2$$

Roots: $x = \tfrac{1}{2}, -2$

✳️ Example 3 (Hard)

Solve $3x^2 – 4x + 5 = 0$

Step 1: $a = 3, b = -4, c = 5$

Step 2: $D = (-4)^2 – 4(3)(5) = 16 – 60 = -44$

Step 3:
Since $D < 0$, roots are imaginary.

$$x = \frac{-(-4) \pm \sqrt{-44}}{2(3)} = \frac{4 \pm i\sqrt{44}}{6}$$

Simplify:

$$x = \frac{2 \pm i\sqrt{11}}{3}$$

✅ Nature of Roots: No real roots (imaginary).

6️⃣ Nature of Roots

Let $D = b^2 – 4ac$

🔹 7️⃣ Summary of Formulas

1. Standard Form: $ax^2 + bx + c = 0$
   

2. Product of roots: $\alpha \beta = \frac{c}{a}$

3. Sum of roots: $\alpha + \beta = -\frac{b}{a}$

4. Quadratic Formula: $x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$

5. Discriminant: $D = b^2 – 4ac$
   

🔹 8️⃣ Practice Questions (Worksheet)

Each question has 3 variants (a, b, c).

Q1. Solve by factorisation:

(a) $x^2 + 7x + 10 = 0$
(b) $x^2 + 9x + 14 = 0$
(c) $x^2 + 11x + 24 = 0$

Q2. Solve by splitting the middle term:

(a) $2x^2 + 7x + 3 = 0$
(b) $3x^2 + 8x + 4 = 0$
(c) $4x^2 + 9x + 2 = 0$

Q3. Find the discriminant and nature of roots:

(a) $x^2 – 4x + 4 = 0$
(b) $x^2 – 6x + 9 = 0$
(c) $x^2 – 10x + 16 = 0$

Q4. Find the sum and product of the roots:

(a) $2x^2 – 5x + 3 = 0$
(b) $3x^2 + 7x + 2 = 0$
(c) $5x^2 – 9x – 2 = 0$