Animated Solution for Mathematics - Quadratic Equations: The sum of the solutions of the equation ∣x−2∣+x(x−4)+2=0,(x>0) is equal to :
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Visualized Solution
The Given Equation
We need to find the sum of solutions for:
∣x−2∣+x(x−4)+2=0
Given condition: x>0
Substitution t=x
Let t=x
Since x>0, the principal square root is positive.
Therefore, t>0
Equation in terms of t
Substitute t into the original equation:
∣t−2∣+t(t−4)+2=0
Expanding the Terms
Expand the term t(t−4):
∣t−2∣+t2−4t+2=0
Completing the Square
Notice the term t2−4t.
We can relate it to the modulus term ∣t−2∣ by completing the square.
Add and subtract 4:
t2−4t+4−4=(t−2)2−4
Simplified Equation in t
Substitute the completed square back:
∣t−2∣+(t−2)2−4+2=0
∣t−2∣+(t−2)2−2=0
Let u=∣t−2∣
Recall the property: (t−2)2=∣t−2∣2
Let u=∣t−2∣
Since u is an absolute value, u≥0
Quadratic Equation in u
Substitute u into the equation:
u+u2−2=0
Rearranging gives:
u2+u−2=0
Factoring the Quadratic
Factorize u2+u−2=0:
u2+2u−u−2=0
u(u+2)−1(u+2)=0
(u−1)(u+2)=0
Solving for u
From the factors, u=1 or u=−2
But we established u≥0
Therefore, we reject u=−2 and keep u=1
Solving for t
Substitute back u=∣t−2∣:
∣t−2∣=1
This gives two cases:
t−2=1⇒t=3
t−2=−1⇒t=1
Solving for x
Recall our first substitution: t=x⇒x=t2
For t=3: x=32=9
For t=1: x=12=1
Final Sum of Solutions
The solutions for x are 1 and 9.
Both satisfy the initial condition x>0.
Sum of solutions =1+9=10
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The Sigma Insight: Solution of Quadratic Equations
Solution Diagram
Analyzing the Setup
The given equation is:
∣x−2∣+x(x−4)+2=0
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 x appears twice. Whenever you see a repeating structure, your first instinct should be to simplify the landscape. Let us introduce a new variable, t=x.
Since the problem specifies x>0, we know that t must also be positive. This simple substitution transforms our intimidating equation into:
∣t−2∣+t(t−4)+2=0
The Elegance of Completing the Square
Expanding the term t(t−4) gives us t2−4t. Now we have:
∣t−2∣+t2−4t+2=0
Look closely at the t2−4t part. It is almost a perfect square. If we add and subtract 4, we get (t2−4t+4)−4, which is (t−2)2−4.
Substituting this back, the equation becomes:
∣t−2∣+(t−2)2−4+2=0
∣t−2∣+(t−2)2−2=0
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 a, a2=∣a∣2. This is a powerful tool in your JEE toolkit.
Let u=∣t−2∣. Since u is a modulus, we must remember the constraint u≥0. Substituting u into our equation, we get:
u+u2−2=0
u2+u−2=0
Factoring this is straightforward:
(u−1)(u+2)=0
This gives us two potential values for u: 1 and −2. But wait! Recall our constraint u≥0. We must reject u=−2 because a modulus cannot be negative. Thus, we are left with u=1.
The Final Reveal
Now, we simply retrace our steps. Since u=∣t−2∣=1, we have two cases:
t−2=1ort−2=−1
This gives us t=3 and t=1. Finally, we return to our original variable x. Since t=x, we have x=t2.
Squaring our values for t, we get:
x=32=9
x=12=1
Both values are positive, satisfying our initial condition. The sum of these solutions is:
1+9=10
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.