Animated Solution for Mathematics - Quadratic Equations: The number of the real roots of the equation $(x+1{)}^2+|x-5|=\frac{27}{4}$ is

Welcome, fellow traveler on the path to JEE mastery. Today, we confront a problem that often intimidates students: an equation involving a modulus.

The equation is $(x+1{)}^2+|x-5|=\frac{27}{4}$. At first glance, the modulus term $|x-5|$ might seem like a barrier, but I want you to see it differently.

Think of the modulus not as a monster, but as a gatekeeper. It has a specific rule: it only cares about the sign of what is inside. If the input is non-negative, it lets it pass through unchanged; if the input is negative, it flips the sign to make it positive.

To work with this gatekeeper, we must find its decision boundary. Where does the expression inside the modulus, $x-5$, change its nature?

It happens exactly at $x=5$. This is our critical point.

It divides the entire real number line into two distinct territories: the region where $x\ge 5$ and the region where $x<5$. By analyzing these two regions separately, we can remove the modulus bars and transform this into a problem of simple algebra.

The High Ground ($x\ge 5$)

Let us first explore the territory where $x\ge 5$. In this region, $x-5$ is always zero or positive. Therefore, the modulus $|x-5|$ simply becomes $x-5$.

Our equation transforms into:

Expanding the square, we get $x^2+2x+1+x-5=\frac{27}{4}$. Combining the like terms, we arrive at $x^2+3x-4=\frac{27}{4}$.

To make our lives easier, let us clear the fraction by multiplying the entire equation by $4$, yielding $4x^2+12x-16=27$, or:

Using the quadratic formula, we find the roots. However, here is the crucial step: we must check if these roots actually live in our territory of $x\ge 5$. Upon calculation, we find that neither root satisfies this condition. Thus, the gatekeeper tells us: no solutions here.

The Low Ground ($x<5$)

Now, let us venture into the region where $x<5$. Here, $x-5$ is negative. To keep the output positive, the modulus must negate the expression, so $|x-5|$ becomes $-(x-5)$, which is $5-x$.

The equation now becomes:

Expanding again, we have $x^2+2x+1+5-x=\frac{27}{4}$, which simplifies beautifully to $x^2+x+6=\frac{27}{4}$. Multiplying by $4$ again, we get $4x^2+4x+24=27$, leading to the quadratic:

This is a much friendlier equation! Solving this using the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, we find two roots: $x=0.5$ and $x=-1.5$. Since both of these values are strictly less than $5$, they are valid, legitimate solutions.

We have traversed both regions. Case 1 yielded no roots, while Case 2 gave us two distinct real roots.

By combining these results, we conclude that the equation has exactly two real roots: $x=0.5$ and $x=-1.5$.

This journey shows that even the most intimidating modulus problems are just a matter of breaking things down into manageable, logical steps. Keep this clarity in your toolkit, and no equation will ever stand in your way again.