SSC CGL Maths Algebra Quadratic Equations: Formulas, Roots, Tricks & Solved Questions

Short and effective solutions for SSC CGL Maths Algebra Quadratic equations help you quickly understand key concepts like roots, coefficients, factorisation, and algebraic identities. Many students struggle with remembering formulas, forming equations from given roots, and solving problems quickly under exam pressure.

Key topics include the relationship between roots and coefficients, the Remainder Theorem, conditions for equal roots, and exam-oriented problem-solving techniques that improve speed and accuracy. Here, we provide you with a clear understanding of SSC CGL Maths Algebra Quadratic Equations to strengthen exam preparation.

Introduction to Quadratic Equations and Roots

A quadratic equation is represented in its general form as:

ax² + bx + c = 0

Here:

• 'a' is the coefficient of x².

• 'b' is the coefficient of x.

• 'c' is the constant term.

The roots of a quadratic equation are the values of x that satisfy the equation. For any quadratic equation, there are two roots, typically represented by alpha (α) and beta (β).

Properties of Roots

For a quadratic equation ax² + bx + c = 0, the following relationships hold:

• Sum of the Roots (α + β):
  α + β = -b/a

• Product of the Roots (α ⋅ β):
  α ⋅ β = c/a

Remember that "sum" has a negative sign in the formula (-b/a), while "product" has a positive sign (c/a).

Problem Solving: Expressions Involving Roots

This section covers calculating expressions using the sum and product of roots.

1. Find 1/α + 1/β

Question: If α and β are the roots of a quadratic equation, find the value of 1/α + 1/β.

Solution:

1. Simplify: 1/α + 1/β = (β + α) / (α ⋅ β)

2. Substitute formulas: (-b/a) / (c/a)

3. Simplify further: -b/c
   Example: For x² + 3x + 7 = 0 (a=1, b=3, c=7), 1/α + 1/β = -3/7.

2. Find α² + β²

Question: Given a quadratic equation, find the value of α² + β².

Solution:

Using the identity (α + β)² = α² + β² + 2αβ, we rearrange to get:

α² + β² = (α + β)² - 2αβ

1. Find sum of roots (α + β) and product of roots (α ⋅ β) from the equation.

2. Substitute these values into the formula.
   Example: For x² - x + 1 = 0, α + β = 1 and α ⋅ β = 1. So, α² + β² = (1)² - 2(1) = -1.

3. Find α/β + β/α

Question: Given a quadratic equation, find the value of α/β + β/α.

Solution:

1. Simplify: α/β + β/α = (α² + β²) / (α ⋅ β)

2. Substitute for α² + β²: ((α + β)² - 2αβ) / (α ⋅ β)

3. Find sum (α + β) and product (α ⋅ β) and substitute.
   Example: For 2x² - 7x + 12 = 0, α + β = 7/2 and α ⋅ β = 6.
   α/β + β/α = ((7/2)² - 2(6)) / 6 = (49/4 - 12) / 6 = (1/4) / 6 = 1/24.

Problem Solving: Finding Unknown Coefficients

1. One Root is Given

Question: If one root of x² - kx + 27 = 0 is 3, find k.

Solution:

Substitute x = 3 into the equation:

(3)² - k(3) + 27 = 0

9 - 3k + 27 = 0

36 - 3k = 0 => 3k = 36 => k = 12.

2. One Root is Twice the Other

Question: Find k if one root of x² - 9x + k = 0 is twice the other.

Solution:

1. Let roots be β and 2β.

2. Sum of roots: β + 2β = -(-9)/1 => 3β = 9 => β = 3.

3. So, roots are 3 and 2(3)=6.

4. Product of roots: (3)(6) = k/1 => k = 18.

3. Sum of Roots Equals Product of Roots

Question: Find k for (k+1)x² + (2k+1)x - k - 5 = 0 if the sum of roots equals the product.

Solution:

1. Condition: α + β = α ⋅ β

2. Substitute formulas: -b/a = c/a => -b = c

3. From equation: a = (k+1), b = (2k+1), c = (-k-5)

4. Substitute into -b = c: -(2k+1) = (-k-5)
   -2k - 1 = -k - 5
   -k = -4 => k = 4.

Forming a Quadratic Equation from its Roots

A quadratic equation can be formed directly from its roots using:

x² - (Sum of Roots)x + (Product of Roots) = 0

Or, x² - (α + β)x + (α ⋅ β) = 0

Derivation: By dividing ax² + bx + c = 0 by 'a', we get x² + (b/a)x + (c/a) = 0. Since - (α + β) = b/a and α ⋅ β = c/a, substitution yields the formula.

1. Form Q.E. Given Sum and Product of Roots

Question: Form the Q.E. where sum of roots is 1 and product is -20.

Solution:

x² - (1)x + (-20) = 0 => x² - x - 20 = 0.

2. Form Q.E. Given the Roots

Question: Which quadratic equation has roots -3 and -5?

Solution:

1. Sum of roots: (-3) + (-5) = -8

2. Product of roots: (-3) ⋅ (-5) = 15

3. Form equation: x² - (-8)x + (15) = 0 => x² + 8x + 15 = 0.

3. Form Q.E. with Reciprocal Roots

Question: Given x² - 5x + 6 = 0, find the Q.E. whose roots are 1/α and 1/β.

Solution:

1. Roots of x² - 5x + 6 = 0 are (x-3)(x-2)=0 => α=3, β=2.

2. New roots: 1/α = 1/3, 1/β = 1/2.

3. New sum: 1/3 + 1/2 = 5/6.

4. New product: (1/3) ⋅ (1/2) = 1/6.

5. New equation: x² - (5/6)x + (1/6) = 0. Multiply by 6: 6x² - 5x + 1 = 0.

4. Form Q.E. with Cubed Roots

Question: If α and β are roots of x² - x + 1 = 0, find the Q.E. whose roots are α³ and β³.

Solution:

1. Original roots: α + β = 1, α ⋅ β = 1.

2. New sum (α³ + β³): Use identity α³ + β³ = (α + β)³ - 3αβ(α + β)
   = (1)³ - 3(1)(1) = 1 - 3 = -2.

3. New product (α³ ⋅ β³): (α ⋅ β)³ = (1)³ = 1.

4. New equation: x² - (-2)x + (1) = 0 => x² + 2x + 1 = 0.

Review of Previous Homework Problem

Question: Given the quadratic equation x² - 3x + 2 = 0, find the quadratic equation whose roots are α+1 and β+1.

Solution:

1. Original roots: α + β = 3, α ⋅ β = 2.

2. New sum: (α+1) + (β+1) = α + β + 2 = 3 + 2 = 5.

3. New product: (α+1)(β+1) = αβ + α + β + 1 = 2 + 3 + 1 = 6.

4. New equation: x² - (5)x + (6) = 0 => x² - 5x + 6 = 0.

Remainder Theorem

The Remainder Theorem states that if a polynomial P(x) is divided by (x - a), the remainder is P(a). If P(x) is completely divisible by (x - a), then P(a) = 0.

1. Find Remainder when Divided by (x-2)

Question: If x⁴ - 3x³ + 2x² - 5x + 7 is divided by (x - 2), what is the remainder?

Solution:

1. Set divisor to zero: x - 2 = 0 => x = 2.

2. Substitute x = 2 into P(x):
   P(2) = (2)⁴ - 3(2)³ + 2(2)² - 5(2) + 7
   = 16 - 24 + 8 - 10 + 7 = -3.

2. Find the remainder when Divided by (3x+2)

Question: If 15x³ + 14x² - 4x + 1 is divided by (3x + 2), what is the remainder?

Solution:

1. Set divisor to zero: 3x + 2 = 0 => x = -2/3.

2. Substitute x = -2/3 into P(x):
   P(-2/3) = 15(-2/3)³ + 14(-2/3)² - 4(-2/3) + 1
   = 15(-8/27) + 14(4/9) + 8/3 + 1
   = -40/9 + 56/9 + 24/9 + 9/9
   = (-40 + 56 + 24 + 9) / 9 = 49/9.

3. Find Remainder when Divided by (x+3)

Question: If 5x³ + 5x² - 6x + 9 is divided by (x + 3), what is the remainder?

Solution:

1. Set divisor to zero: x + 3 = 0 => x = -3.

2. Substitute x = -3 into P(x):
   P(-3) = 5(-3)³ + 5(-3)² - 6(-3) + 9
   = 5(-27) + 5(9) + 18 + 9
   = -135 + 45 + 18 + 9 = -63.

Finding Roots by Factoring

1. Finding Roots of x² - 8x + 15 = 0

Solution:

1. Find two numbers whose product is 15 and sum is -8. These are -3 and -5.

2. Break the middle term: x² - 3x - 5x + 15 = 0

3. Factor by grouping: x(x - 3) - 5(x - 3) = 0
   (x - 3)(x - 5) = 0

4. Set each factor to zero: x - 3 = 0 => x = 3; x - 5 = 0 => x = 5.

When factoring x² + Bx + C = 0, find two numbers (p, q) such that p+q = B and p*q = C. The factors are (x+p) and (x+q), and the roots are -p and -q (change the sign).

Condition for Equal Roots

For a quadratic equation ax² + bx + c = 0, the condition for having equal roots is that its discriminant (D) must be zero.

D = b² - 4ac = 0

This implies: b² = 4ac.

1. Find k for Equal Roots

Question: Find the positive value of k for which 9x² + 3kx + 4 = 0 has equal roots.

Solution:

1. Identify a=9, b=3k, c=4.

2. Apply b² = 4ac: (3k)² = 4(9)(4)
   9k² = 144
   k² = 16

3. Solve for k: k = ±4. Since a positive value is asked, k = 4.

2. Express c in terms of a and b for Equal Roots

Question: If ax² + bx + c = 0 has equal roots, express c in terms of a and b.

Solution:

1. Apply b² = 4ac.

2. Solve for c: c = b² / 4a.

Combined Roots and Product Problem

1. Evaluate 1/α + 1/β + 2αβ

Question: Given x² - 5x + 4 = 0, evaluate 1/α + 1/β + 2αβ.

Solution:

1. Simplify: 1/α + 1/β + 2αβ = (α + β) / (α ⋅ β) + 2αβ.

2. From x² - 5x + 4 = 0: α + β = 5, α ⋅ β = 4.

3. Substitute: (5) / (4) + 2(4) = 5/4 + 8 = 5/4 + 32/4 = 37/4.

Algebra: Value Putting Techniques

These techniques simplify complex algebraic expressions by substituting specific values, especially useful in competitive exams.

1. Polynomial Evaluation with (x+1) Coefficients

This technique applies when coefficients in a polynomial are related to x (e.g., x+1 or x-1).

Problem: Evaluate x⁵ - 12x⁴ + 12x³ - 12x² + 12x - 5, given x = 11.

Solution:

Observe that 12 = x+1. Substitute (x+1) for 12:

x⁵ - (x+1)x⁴ + (x+1)x³ - (x+1)x² + (x+1)x - 5

= x⁵ - x⁵ - x⁴ + x⁴ + x³ - x³ - x² + x² + x - 5

= x - 5

Substitute x = 11: 11 - 5 = 6.

Problem: Evaluate x⁴ - 9x³ + 9x² - 9x - 8, given x = 8.

Solution:

Observe that 9 = x+1. Substitute (x+1) for 9:

x⁴ - (x+1)x³ + (x+1)x² - (x+1)x - 8

= x⁴ - x⁴ - x³ + x³ + x² - x² - x - 8

= -x - 8

Substitute x = 8: -8 - 8 = -16.

2. Equal Coefficients Condition

Condition: If
a² + b² + c² = ab + bc + ca,
then it implies that
a = b = c (for real numbers).

Correct Reasoning:

Start by rearranging the given condition:

a² + b² + c² − ab − bc − ca = 0

Now use the identity:

(a − b)² + (b − c)² + (c − a)² = 2(a² + b² + c² − ab − bc − ca)

So we get:

(a − b)² + (b − c)² + (c − a)² = 0

Each term in the expression is a square, and therefore always greater than or equal to zero. The only way their sum can be zero is when each term individually equals zero:

a − b = 0
b − c = 0
c − a = 0

Hence,

a = b = c

Conclusion:
The given condition is satisfied only when all three variables are equal, which makes it useful in simplifying symmetric expressions in algebra.

3. Breaking Down Constants in Equations

This technique involves equating each term to a common constant when a sum of terms equals that constant.

Problem: Solve for x in (x - a²/(b²+c²)) + (x - b²/(c²+a²)) + (x - c²/(a²+b²)) = 3.

Solution:

Since the sum is 3, we can assume each term equals 1:

x - a²/(b²+c²) = 1

x = 1 + a²/(b²+c²)

x = (b²+c² + a²)/(b²+c²)

This leads to consistent expressions involving a², b², and c², allowing simplification to a single symmetric form.

This logic holds for all three terms, giving a consistent value for x.

4. Value Putting with Symmetrical Equations

For symmetrical equations, assuming equal values for variables simplifies calculations significantly.

Problem: Given a² = b+c, b² = c+a, c² = a+b. Find the value of 1 / (a + 1) + 1 / (b + 1) + 1 / (c + 1).

Solution:

Assume a = b = c. Substitute into a² = b+c:

a² = a+a => a² = 2a

a² - 2a = 0 => a(a - 2) = 0

So, a = 0 or a = 2.

If a=0, then a,b,c=0, which might lead to division by zero.

Choose a = 2. Thus, a = b = c = 2.

Check: 2² = 2+2 => 4 = 4 (Satisfied).

Substitute a=b=c=2 into the expression:

1/(2+1) + 1/(2+1) + 1/(2+1) = 1/3 + 1/3 + 1/3 = 1.

5. Problem Given a + b + c = 0, find the value of
a² / (a² − bc) + b² / (b² − ca) + c² / (c² − ab)

Solution:

From the given condition:
a + b + c = 0 ⇒ c = −(a + b)

Step 1: Simplify denominators

Substitute c = −(a + b):

• a² − bc = a² − b(−a − b) = a² + ab + b²

• b² − ca = b² − a(−a − b) = b² + ab + a² = a² + b² + ab

• c² − ab = (a + b)² − ab = a² + 2ab + b² − ab = a² + b² + ab

So all denominators become:

a² + b² + ab

Step 2: Rewrite the expression

= a² / (a² + b² + ab) + b² / (a² + b² + ab) + c² / (a² + b² + ab)

= (a² + b² + c²) / (a² + b² + ab)

Step 3: Simplify the numerator

Since c = −(a + b):

c² = (a + b)² = a² + b² + 2ab

So,

a² + b² + c²
= a² + b² + (a² + b² + 2ab)
= 2a² + 2b² + 2ab
= 2(a² + b² + ab)

Step 4: Final value

= 2(a² + b² + ab) / (a² + b² + ab)

= 2

6. Finding a³ + b³ with Given Sum and Product

Problem: Given a + b = 6 and ab = 5. Find a³ + b³.

Solution:

Find two numbers whose product is 5 and sum is 6. These are 5 and 1.

So, let a = 5 and b = 1.

Substitute into a³ + b³:

5³ + 1³ = 125 + 1 = 126.

7. Finding ab with a given sum of Cubes and Sum

Problem: Given a³ + b³ = 217 and a + b = 7. Find ab.

Solution:

Observe a³ + b³ = 217. The cube of 6 is 216 (6³ = 216).

If one number is 6, then 6³ = 216. For the sum to be 217, the other number's cube must be 1 (1³ = 1).

So, let a = 6 and b = 1.

Check a + b = 7: 6 + 1 = 7 (satisfied).

Find ab: 6 * 1 = 6.