Sigma Percentile
JEE Main 2019 (8 April Shift 1)
LEVELJEE Main

Animated Solution for Mathematics - Quadratic Equations: The sum of the solutions of the equation is equal to :

Select Answer:

Visualized Solution

The Given Equation

  • We need to find the sum of solutions for:
  • Given condition:

Substitution

  • Let
  • Since , the principal square root is positive.
  • Therefore,

Equation in terms of

  • Substitute into the original equation:

Expanding the Terms

  • Expand the term :

Completing the Square

  • Notice the term .
  • We can relate it to the modulus term by completing the square.
  • Add and subtract :

Simplified Equation in

  • Substitute the completed square back:

Let

  • Recall the property:
  • Let
  • Since is an absolute value,

Quadratic Equation in

  • Substitute into the equation:
  • Rearranging gives:

Factoring the Quadratic

  • Factorize :

Solving for

  • From the factors, or
  • But we established
  • Therefore, we reject and keep

Solving for

  • Substitute back :
  • This gives two cases:

Solving for

  • Recall our first substitution:
  • For :
  • For :

Final Sum of Solutions

  • The solutions for are and .
  • Both satisfy the initial condition .
  • Sum of solutions

The Sigma Insight: Solution of Quadratic Equations

Solution Diagram

Analyzing the Setup

The given equation is:
At first glance, it looks like a chaotic mess of radicals and absolute values. However, in the world of JEE Advanced, chaos is just a hidden order waiting to be discovered.
Notice how appears twice. Whenever you see a repeating structure, your first instinct should be to simplify the landscape. Let us introduce a new variable, .
Since the problem specifies , we know that must also be positive. This simple substitution transforms our intimidating equation into:

The Elegance of Completing the Square

Expanding the term gives us . Now we have:
Look closely at the part. It is almost a perfect square. If we add and subtract , we get , which is .
Substituting this back, the equation becomes:
This is the moment where the problem shifts from a struggle to a dance. We have successfully aligned the terms to match the modulus.

The Power of Substitution

We know that for any real number , . This is a powerful tool in your JEE toolkit.
Let . Since is a modulus, we must remember the constraint . Substituting into our equation, we get:
Factoring this is straightforward:
This gives us two potential values for : and . But wait! Recall our constraint . We must reject because a modulus cannot be negative. Thus, we are left with .

The Final Reveal

Now, we simply retrace our steps. Since , we have two cases:
This gives us and . Finally, we return to our original variable . Since , we have .
Squaring our values for , we get:
Both values are positive, satisfying our initial condition. The sum of these solutions is:
And there you have it—a complex problem dismantled by the power of substitution and logical consistency. Keep practicing this mindset, and you will find that even the toughest problems have a beautiful, logical soul.

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