Sigma Percentile
JEE Main 2021 (25 July Shift 2)
LEVELBoard

Animated Solution for Mathematics - Quadratic Equations: The number of real solutions of the equation, is:

Select Answer:

Visualized Solution

The Equation

  • Given equation:

Key Property

  • Recall that for any real number ,

Substitution

  • Let

Constraint on

  • Since , and absolute value is non-negative:

Transformed Equation

  • Substituting , the equation becomes:

Splitting the Middle Term

  • Split the middle term into :

Factorization

  • Take common factors:

Roots of

  • Equating each factor to zero:

Checking Constraints

  • We know
  • Therefore, is rejected.
  • Valid value:

Back-Substitution

  • Substitute back into :

Final Solutions

  • Removing the absolute value gives:

Graphical View

  • The graph of intersects the x-axis at exactly two points: and
  • Total real solutions =

The Sigma Insight: Solution of Quadratic Equations

Solution Diagram

Analyzing the Setup

The equation provided is . While it resembles a standard quadratic equation, the presence of the absolute value term introduces a specific constraint that we must address carefully.

The Hidden Symmetry

The first step is to recognize the relationship between and . For any real number , the square of the number is identical to the square of its absolute value:
By substituting this identity into our original equation, we transform it into a quadratic form involving only the absolute value:

The Bridge of Substitution

To simplify the algebra, let us introduce a substitution variable. Let .
Our equation now becomes a standard quadratic:
However, we must respect the gatekeeper constraint. Since and the absolute value of any real number is non-negative, we must enforce the condition:

Solving the Quadratic

We now factor the quadratic equation . We look for two numbers that multiply to and add to . These numbers are and .
The factored form is:
This yields two potential roots for :

Final Calculation

We must now check these roots against our constraint . The value is rejected because an absolute value cannot be negative.
This leaves us with the single valid solution:
Substituting back for , we have . This implies that can take two possible values:
or

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