Theory & Concepts

SSC CGL Quadratic Equations Questions, Formulas & Tricks

Get comprehensive theory, expert shortcuts, and hand-picked practice questions for Quadratic Equations specifically designed for the SSC CGL 2025-26 pattern.

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25 min readDifficulty: Intermediate

Quadratic equations are central to the Advanced Maths section of SSC CGL. Beyond simply finding roots via factorization, examiners love testing the conceptual properties of roots (sum, product, and nature) without requiring you to actually solve the equation.

Learning path

Splitting the Middle Term
The Discriminant Rule
Sum & Product of Roots
10 Exam-Level Examples

1. Nature of Roots (Discriminant)

For a standard quadratic equation $ax^2 + bx + c = 0$ , the Discriminant $(D = b^2 - 4ac)$ tells you exactly what kind of roots the equation has without solving it.

Real Roots

If $D > 0$ : Roots are Real and Unequal.

If $D = 0$ : Roots are Real and Equal.

Note: SSC frequently asks to "find k for which roots are equal". Simply set $b^2 - 4ac = 0$ .

Imaginary Roots

If $D < 0$ : Roots are Imaginary (Complex Conjugates).

The graph of this equation never touches the x-axis.

2. Sum & Product of Roots

If $\alpha$ and $\beta$ are the roots of the equation $ax^2 + bx + c = 0$ , you can instantly find their sum and product.

Core Property

You do not need to solve the equation to find these:

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

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

Be very careful with the negative sign in the Sum formula. If the equation is $x^2 - 5x + 6 = 0$ , the sum is $-(-5)/1 = 5$ .

3. Constructing an Equation

If you are given the roots, or the sum and product of the roots, you can instantly build the quadratic equation.

The Builder Formula

x^2 - (\text{Sum})x + (\text{Product}) = 0

Always remember the negative sign is before the Sum.

Reciprocal Roots Hack

To find an equation whose roots are reciprocals of $ax^2 + bx + c = 0$ :

Swap $a$ and $c$ : $cx^2 + bx + a = 0$ .

4. 10 Solved examples

Question 01Exam Pattern

Find the roots of the equation $ x^2 - 5x + 6 = 0 $.

2, 3

-2, -3

1, 6

-1, -6

Correct answer: a) 2, 3

Solution

Step 1: We need two numbers that multiply to 6 and add to -5.

Step 2: The numbers are -2 and -3.

Step 3: Factorize the equation: $ (x - 2)(x - 3) = 0 $.

Final calculation: Setting each factor to zero gives $ x = 2 $ and $ x = 3 $.

Question 02Exam Pattern

What is the nature of the roots of $ 2x^2 - 4x + 3 = 0 $?

Real and Equal

Real and Distinct

Imaginary

Rational

Correct answer: c) Imaginary

Solution

Step 1: Calculate the discriminant $ D = b^2 - 4ac $.

Step 2: Here, $ a = 2, b = -4, c = 3 $.

Step 3: $ D = (-4)^2 - 4(2)(3) = 16 - 24 = -8 $.

Final calculation: Since $ D < 0 $, the roots are imaginary.

Question 03Exam Pattern

Find the value of $ k $ for which the roots of $ x^2 - kx + 9 = 0 $ are real and equal.

-3

±6

±9

Correct answer: c) ±6

Solution

Step 1: For real and equal roots, the discriminant must be zero: $ D = 0 $.

Step 2: $ D = (-k)^2 - 4(1)(9) = 0 $.

Step 3: $ k^2 - 36 = 0 $.

Final calculation: $ k^2 = 36 \implies k = \pm 6 $.

Question 04Exam Pattern

If the sum of the roots of $ kx^2 + 2x + 3k = 0 $ is equal to their product, find $ k $.

-2/3

2/3

3/2

-3/2

Correct answer: a) -2/3

Solution

Step 1: Sum of roots $ = -b/a = -2/k $.

Step 2: Product of roots $ = c/a = 3k/k = 3 $.

Step 3: Set them equal: $ -2/k = 3 $.

Final calculation: Cross multiply to get $ 3k = -2 \implies k = -2/3 $.

Question 05Exam Pattern

Form a quadratic equation whose roots are 4 and -5.

$ x^2 + x - 20 = 0 $

$ x^2 - x - 20 = 0 $

$ x^2 + 9x - 20 = 0 $

$ x^2 - 9x - 20 = 0 $

Correct answer: a) $ x^2 + x - 20 = 0 $

Solution

Step 1: Calculate the Sum of the roots: $ 4 + (-5) = -1 $.

Step 2: Calculate the Product of the roots: $ 4 \times -5 = -20 $.

Step 3: Use the formula: $ x^2 - (\text{Sum})x + (\text{Product}) = 0 $.

Final calculation: $ x^2 - (-1)x + (-20) = 0 \implies x^2 + x - 20 = 0 $.

Question 06Exam Pattern

If $ \alpha $ and $ \beta $ are the roots of $ 2x^2 - 5x + 3 = 0 $, find the value of $ \alpha^2 + \beta^2 $.

13/4

15/4

25/4

19/4

Correct answer: a) 13/4

Solution

Step 1: Find Sum $ (\alpha + \beta) = -(-5)/2 = 5/2 $.

Step 2: Find Product $ (\alpha\beta) = 3/2 $.

Step 3: Use the algebraic identity: $ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta $.

Step 4: Substitute the values: $ (5/2)^2 - 2(3/2) $.

Final calculation: $ 25/4 - 3 = (25 - 12)/4 = 13/4 $.

Question 07Exam Pattern

The product of two consecutive positive integers is 156. Find the integers.

11, 12

12, 13

13, 14

14, 15

Correct answer: b) 12, 13

Solution

Step 1: Let the integers be $ x $ and $ x+1 $.

Step 2: The equation is $ x(x+1) = 156 \implies x^2 + x - 156 = 0 $.

Step 3: Find two numbers that multiply to -156 and add to 1. They are 13 and -12.

Step 4: Factorize: $ (x+13)(x-12) = 0 $.

Final calculation: Since the integers must be positive, $ x = 12 $. The next integer is 13.

Question 08Exam Pattern

If one root of the equation $ x^2 - 8x + k = 0 $ is 3 times the other, find $ k $.

Correct answer: a) 12

Solution

Step 1: Let the roots be $ a $ and $ 3a $.

Step 2: Sum of roots $ = a + 3a = 4a $. From equation, Sum $ = -(-8)/1 = 8 $.

Step 3: $ 4a = 8 \implies a = 2 $. So the roots are 2 and 6.

Step 4: Product of roots $ = c/a = k $.

Final calculation: Product is $ 2 \times 6 = 12 \implies k = 12 $.

Question 09Exam Pattern

Find the equation whose roots are the reciprocals of the roots of $ 3x^2 - 7x + 2 = 0 $.

$ 2x^2 + 7x + 3 = 0 $

$ 2x^2 - 7x + 3 = 0 $

$ 3x^2 + 7x + 2 = 0 $

$ 2x^2 - 7x - 3 = 0 $

Correct answer: b) $ 2x^2 - 7x + 3 = 0 $

Solution

Step 1: Use the Reciprocal Roots hack. Swap the coefficient of $ x^2 $ (a) and the constant term (c).

Step 2: Here $ a = 3 $ and $ c = 2 $.

Final calculation: The new equation becomes $ 2x^2 - 7x + 3 = 0 $.

Question 010Exam Pattern

Solve for $ x $: $ x - \frac{1}{x} = \frac{3}{2} $.

2, -1/2

1/2, -2

3, -1/3

1/3, -3

Correct answer: a) 2, -1/2

Solution

Step 1: Multiply the entire equation by $ 2x $ to remove the denominators.

Step 2: $ 2x^2 - 2 = 3x $.

Step 3: Rearrange into standard form: $ 2x^2 - 3x - 2 = 0 $.

Step 4: Multiply to -4, add to -3. The numbers are -4 and 1.

Step 5: Split middle term: $ 2x^2 - 4x + x - 2 = 0 \implies 2x(x-2) + 1(x-2) = 0 $.

Final calculation: $ (2x+1)(x-2) = 0 \implies x = 2 \text{ or } x = -1/2 $.

SSC CGL Quadratic Equations Questions, Formulas & Tricks

Learning path

1. Nature of Roots (Discriminant)

Real Roots

Imaginary Roots

2. Sum & Product of Roots

3. Constructing an Equation

The Builder Formula

Reciprocal Roots Hack

4. 10 Solved examples

Find the roots of the equation \( x^2 - 5x + 6 = 0 \).

What is the nature of the roots of \( 2x^2 - 4x + 3 = 0 \)?

Find the value of \( k \) for which the roots of \( x^2 - kx + 9 = 0 \) are real and equal.

If the sum of the roots of \( kx^2 + 2x + 3k = 0 \) is equal to their product, find \( k \).

Form a quadratic equation whose roots are 4 and -5.

If \( \alpha \) and \( \beta \) are the roots of \( 2x^2 - 5x + 3 = 0 \), find the value of \( \alpha^2 + \beta^2 \).

The product of two consecutive positive integers is 156. Find the integers.

If one root of the equation \( x^2 - 8x + k = 0 \) is 3 times the other, find \( k \).

Find the equation whose roots are the reciprocals of the roots of \( 3x^2 - 7x + 2 = 0 \).

Solve for \( x \): \( x - \frac{1}{x} = \frac{3}{2} \).