JEE Main · Mathematics ↑ Rising

Quadratic Equations appeared 24 times across 3 years — 2.8% of Mathematics. This question is from Equations Involving Modulus.

Year 2026 2025 2024 Total
Questions 9 10 5 24

The sum, of the squares of all the roots of the equation x² + |2x - 3| - 4 = 0, is:

Solution & Explanation

Related Formula

Modulus definition rule:

|x| = cases x, & x ≥ 0 -x, & x < 0 cases
Core Logic

Analyze the roots by splitting into cases around the critical threshold x = (3)/(2):

Case I: x ≥ (3)/(2)

x² + 2x - 3 - 4 = 0 x² + 2x - 7 = 0 x = 2√(2) - 1

(We select the positive root since 2√(2)-1 ≥ 1.5).

Step 1: Evaluating the alternate domain branch

Case II: x < (3)/(2)

x² - (2x - 3) - 4 = 0 x² - 2x - 1 = 0 x = 1 - √(2)

(We select 1-√(2) since it satisfies the inequality constraint).

Step 2: Summing the Squares of the Roots
Sum of Squares = (2√(2) - 1)² + (1 - √(2))² = (8 - 4√(2) + 1) + (1 - 2√(2) + 2) = 12 - 6√(2) = 6(2 - √(2))
Pattern Recognition

Always validate absolute root values against their domain restrictions to avoid including phantom solutions.

Chapter Mix

Class 11 Maths: Quadratic Equations

More Quadratic Equations Previous-Year Questions — Page 5

Q28 jee_main_2024_30_january_evening Modulus Equations
The number of real solutions of the equation x(x² + 3|x| + 5|x - 1| + 6|x - 2|) = 0 is
Numerical Answer. Answer: 1 to 1

Solution

Related Formula
Zero Product Property: A · B = 0 A = 0 or B = 0
Core Logic

Given equation:

x(x² + 3|x| + 5|x - 1| + 6|x - 2|) = 0

This factors into two possibilities:

  • x = 0
  • x² + 3|x| + 5|x - 1| + 6|x - 2| = 0
Step 1: Evaluating the Modulus Term

Look at the second factor: f(x) = x² + 3|x| + 5|x - 1| + 6|x - 2|. Notice that all terms inside are strictly non-negative:

  • x² ≥ 0
  • 3|x| ≥ 0
  • 5|x - 1| ≥ 0
  • 6|x - 2| ≥ 0
  • For the sum to be 0, every single term must be simultaneously zero. x² = 0 x = 0 However, if x = 0, then 5|x-1| = 5(1) = 5 ≠ 0. Therefore, there is no real value of x that makes this entire second factor equal to zero.

Step 2: Conclusion

The only valid solution to the equation is x = 0 from the first factor. Thus, there is exactly 1 real solution.

Pattern Recognition

A sum of absolute values and squares set to 0 requires all individual components to hit 0 concurrently. If they have different zero-nodes (0, 1, 2), the sum can never be 0.

Chapter Mix

Class 11 Maths: Quadratic Equations

Q22 jee_main_2024_31_jan_evening Roots of Quadratic Equation
Let a, b, c be the length of three sides of a triangle satisfying the condition (a² + b²)x² - 2b(a + c)x + (b² + c²) = 0. If the set of all possible values of x is the interval (α, β) then 12(α² + β²) is equal to
Numerical Answer. Answer: 36 to 36

Solution

Core Logic

Given equation: (a²+b²)x² - 2b(a+c)x + b²+c² = 0. Expand and rearrange into perfect squares:

(a²x² - 2abx + b²) + (b²x² - 2bcx + c²) = 0 (ax - b)² + (bx - c)² = 0

Since squares must be non-negative, each term is zero:

ax - b = 0 x = (b)/(a) bx - c = 0 x = (c)/(b)

Thus, b = ax and c = bx = ax². Since a,b,c form a triangle, the triangle inequality holds:

  • a + b > c a + ax > ax² x² - x - 1 < 0 1-√(5)2 < x < 1+√(5)2
  • a + c > b a + ax² > ax x² - x + 1 > 0 (Always true for real x)
  • b + c > a ax + ax² > a x² + x - 1 > 0 x > -1+√(5)2 or x < -1-√(5)2.
  • Taking the intersection (and noting x > 0 since sides are positive):

√(5)-12 < x < √(5)+12

So, α = √(5)-12 and β = √(5)+12. Calculate 12(α² + β²):

12 ( 6-2√(5)4 + 6+2√(5)4 ) = 12 ( (12)/(4) ) = 36
Chapter Mix

Class 11 Maths: Complex Numbers and Quadratic Equations Class 11 Maths: Straight Lines

Q1 jee_main_2024_31_jan_morning Nature of Roots
For 0 < c < b < a, let (a + b - 2c)x² + (b + c - 2a)x + (c + a - 2b) = 0 and α ≠ 1 be one of its root. Then, among the two statements (I) If α in (-1,0), then b cannot be the geometric mean of a and c (II) If α in (0,1), then b may be the geometric mean of a and c
  • A. Both (I) and (II) are true
  • B. Neither (I) nor (II) is true
  • C. Only (II) is true
  • D. Only (I) is true

Solution

Related Formula
Sum of coefficients = 0 x = 1 is a root.
Core Logic

Given f(x) = (a + b - 2c)x² + (b + c - 2a)x + (c + a - 2b) = 0. Substituting x = 1:

f(1) = a + b - 2c + b + c - 2a + c + a - 2b = 0

Thus, one root is 1. Let the other root be α.

Step 1: Find the other root

Product of roots = (c + a - 2b)/(a + b - 2c). Since one root is 1, we have:

α · 1 = (c + a - 2b)/(a + b - 2c) α = (c + a - 2b)/(a + b - 2c)
Step 2: Analyze Statement (I)

If -1 < α < 0:

-1 < (c + a - 2b)/(a + b - 2c) < 0

This implies b > (a + c)/(2) and b + c < 2a. Therefore, b cannot be the Geometric Mean of a and c. Statement (I) is true.

Step 3: Analyze Statement (II)

If 0 < α < 1:

0 < (c + a - 2b)/(a + b - 2c) < 1

This gives b > c and b < (a + c)/(2). Therefore, b may be the Geometric Mean between a and c. Statement (II) is true.

Pattern Recognition

When coefficients in a quadratic equation are cyclic and sum to 0, one root is always 1. The other root is directly c/a.

Chapter Mix

Class 11 Maths: Quadratic Equations Class 11 Maths: Sequences and Series

Q20 jee_main_2024_31_jan_morning Sign of Quadratic Expressions
Let S be the set of positive integral values of a for which (ax² + 2(a + 1)x + 9a + 4)/(x² - 8x + 32) < 0, x in R. Then, the number of elements in S is:
  • A. 1
  • B. 0
  • C. ∞
  • D. 3

Solution

Core Logic

For the denominator x² - 8x + 32, D = 64 - 128 < 0 and a = 1 > 0. Thus, x² - 8x + 32 > 0 x in R.

Step 1: Constraint on Numerator

Since the denominator is always positive, the numerator must be strictly negative for all x in R.

ax² + 2(a + 1)x + 9a + 4 < 0 x in R

This requires a < 0 and D < 0.

Step 2: Conclusion

Since a must be strictly less than 0, there are no positive integral values of a that satisfy the condition. Hence, S is an empty set. Number of elements is 0.

Chapter Mix

Class 11 Maths: Quadratic Equations

Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)