Animated Solution for Mathematics - Quadratic Equations: The number of real roots of the equation $x|x|-5|x+2|+6=0$, is

Unmasking the Equation

Welcome, fellow travelers on the JEE journey! Today, we are going to dissect a problem that, at first glance, might look like a simple algebraic equation, but is actually a beautiful exercise in logical partitioning.

We are looking at the equation:

The modulus function is the great shape-shifter of mathematics. It is a filter that ensures everything it touches comes out non-negative. But to solve an equation containing it, we must strip away that mask.

We do this by identifying the critical points. Look at the terms $|x|$ and $|x+2|$. The first changes its nature at $x=0$, and the second at $x=-2$. These two points are the anchors of our analysis, dividing the real number line into three distinct zones.

The Negative Zone ($x<-2$)

Imagine standing on the number line to the left of $-2$. If you pick any number here, say $-3$, both $x$ and $x+2$ are negative.

Because the modulus function negates negative inputs to make them positive, we replace $|x|$ with $-x$ and $|x+2|$ with $-(x+2)$. Our equation transforms into:

Simplifying this, we get $-x^2+5x+10+6=0$, which rearranges to $x^2-5x-16=0$. Using the quadratic formula, we find potential roots at:

Now, here is the crucial step: we must check if these roots live in our zone. $\sqrt{89}$ is roughly $9.43$. The root $x=\frac{5+9.43}{2}\approx 7.21$ is clearly not less than $-2$, so we discard it. However, $x=\frac{5-9.43}{2}\approx -2.215$ fits perfectly. That is our first victory!

The Transition Zone ($-2\le x<0$)

Now, we step into the middle ground. Here, $x$ is still negative, so $|x|$ remains $-x$. But $x+2$ is now positive (e.g., $-1+2=1$).

Thus, $|x+2|$ simply becomes $x+2$. Our equation shifts again:

This simplifies to $-x^2-5x-4=0$, or $x^2+5x+4=0$. Factoring this is a joy:

We get $x=-1$ and $x=-4$. Again, we check our contract. Our zone is $[-2,0)$. The value $x=-1$ is inside, but $x=-4$ is outside. We keep $-1$ and reject $-4$. That is our second valid root!

The Positive Zone ($x\ge 0$)

Finally, we reach the land of the positive. Here, both $x$ and $x+2$ are non-negative. The modulus functions vanish, leaving us with:

This simplifies to $x^2-5x-4=0$. Applying the quadratic formula, we get:

With $\sqrt{41}\approx 6.4$, the root $x=\frac{5+6.4}{2}\approx 5.7$ is valid because it is $\ge 0$. The other root, $x=\frac{5-6.4}{2}\approx -0.7$, is rejected.

We have traversed all three regions. We found one valid root in each:

Counting them up, we have exactly three real roots. This problem wasn't just about solving a quadratic; it was about respecting the boundaries of the modulus function. Keep this discipline, and no equation will ever be able to hide its roots from you!