Animated Solution for Mathematics - Quadratic Equations: The number of real solutions of the equation $x|x+5|+2|x+7|-2=0$ is_________

Welcome, fellow traveler on the road to JEE Advanced. Today, we are going to dismantle a problem that often strikes fear into the hearts of students: the modulus equation.

Specifically, we are looking at $x|x+5|+2|x+7|-2=0$. Many students see those vertical bars and immediately freeze, thinking they need a special formula or a magic trick.

But here is the secret: the modulus is not a monster. It is simply a sign-shifter. It is a piecewise function that asks, 'Am I positive or negative?' and acts accordingly. Let us embark on this journey to solve it step-by-step.

The first step in any modulus problem is to identify the 'hinge points'—the values of $x$ where the expression inside the modulus changes from positive to negative. Look at the terms $|x+5|$ and $|x+7|$.

When does $x+5$ become zero? At $x=-5$. When does $x+7$ become zero? At $x=-7$.

These are our boundaries. Think of them as the pillars of a bridge. They divide the entire real number line into three distinct zones: the region to the right of $-5$, the region between $-7$ and $-5$, and the region to the left of $-7$.

By splitting the problem into these three zones, we turn one scary equation into three simple, manageable ones.

Imagine you are standing to the right of $-5$. Pick any number, say $0$. If you plug $0$ into $x+5$, you get $5$, which is positive. If you plug it into $x+7$, you get $7$, which is also positive.

Because the expressions inside the modulus are positive, the modulus signs are effectively invisible. They do nothing. So, our equation simplifies beautifully to:

Expanding this, we get $x^2+5x+2x+14-2=0$, which simplifies to $x^2+7x+12=0$. This is a classic quadratic!

Factoring it gives us $(x+3)(x+4)=0$, yielding roots $x=-3$ and $x=-4$. But wait—we must check our contract. Did we assume $x\ge -5$? Yes.

Are $-3$ and $-4$ both greater than or equal to $-5$? Yes, they are. We have found two valid solutions!

Now, let us step into the middle zone. Pick a number like $-6$. If you plug $-6$ into $x+5$, you get $-1$ (negative). But if you plug it into $x+7$, you get $1$ (positive).

This is the crucial moment. The first modulus must flip its sign to become positive, so $|x+5|$ becomes $-(x+5)$. The second modulus remains as it is, $x+7$.

Our equation transforms into:

Expanding this, we get $-x^2-5x+2x+14-2=0$, which simplifies to $-x^2-3x+12=0$. Multiplying by $-1$ gives $x^2+3x-12=0$.

This does not factorize cleanly, so we use the quadratic formula:

Since $\sqrt{57}$ is approximately $7.55$, our roots are roughly $2.27$ and $-5.27$. We must check our contract: $-7\le x<-5$.

The root $2.27$ is clearly out of bounds. However, $-5.27$ fits perfectly! We have found our third solution.

Finally, we look to the left of $-7$. Pick $-10$. Both $x+5$ and $x+7$ are negative. Therefore, both modulus terms must flip their signs.

The equation becomes:

Expanding this, we get $-x^2-5x-2x-14-2=0$, which simplifies to $-x^2-7x-16=0$, or $x^2+7x+16=0$.

Let us check the discriminant: $D=7^2-4(1)(16)=49-64=-15$. Since the discriminant is negative, there are no real roots here.

We have traversed all three territories. We found two solutions in the first, one in the second, and none in the third.

Adding them up, we have exactly three real solutions: $x=-3,x=-4,$ and $x=\frac{-3-\sqrt{57}}{2}$.

You see? The modulus is not a barrier; it is just a path. By breaking the problem down, you conquered it. Keep this confidence for your next challenge!