The sum, of the squares of all the roots of the equation x^2 + |2x - 3| - 4 = 0 is

Q: Why do we have to reject some of the roots we found?

Great question! When we solve a modulus equation by cases, we are essentially saying, 'If this condition is true, then this equation holds.' If we find a root that violates the condition we started with (e.g., we assumed $x \ge 1.5$ but found a root like $-3.82$), that root is mathematically inconsistent with the assumption that allowed us to remove the modulus bars in the first place. Therefore, it is not a solution to the original equation.

Q: Can I just square both sides of the equation to get rid of the modulus?

Squaring both sides of an equation involving a modulus is a dangerous path. While it can sometimes work, it often introduces extraneous roots and makes the algebra significantly more complex, leading to a quartic equation ($x^4$) that is much harder to solve. The case-based approach is the standard, most reliable method for JEE Advanced problems involving modulus functions.

Q: Is there a graphical way to solve this?

Absolutely! You could plot $y = x^2$ and $y = 4 - |2x - 3|$ on the same graph. The intersection points of these two curves would give you the roots. While this is a fantastic way to visualize the problem and verify your algebraic results, the algebraic method is usually required for the precise calculation of the sum of squares.

Question

The sum, of the squares of all the roots of the equation $x^{2}+|2x-3|-4=0$ is

Options:

(A) $3(3-\sqrt{2})$

(B) $6(3-\sqrt{2})$

(C) $6(2-\sqrt{2})$

(D) $3(2-\sqrt{2})$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The equation provided is $x^2 + |2x - 3| - 4 = 0$. The modulus term $|2x - 3|$ acts as a gatekeeper, changing its behavior based on the sign of the expression inside.

To determine the critical point where the behavior shifts, we set the internal expression to zero:
$$2x - 3 = 0 \Rightarrow x = 1.5$$

This point divides the number line into two distinct regions. We must analyze each region separately to find the valid solutions.

### Case 1: The Right Side of the Boundary

We consider the region where $x \ge 1.5$. In this territory, the expression $2x - 3$ is non-negative, allowing us to remove the modulus bars directly:
$$x^2 + (2x - 3) - 4 = 0$$

Simplifying this expression leads to the standard quadratic equation:
$$x^2 + 2x - 7 = 0$$

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we calculate the roots:
$$x = \frac{-2 \pm \sqrt{4 - 4(1)(-7)}}{2} = \frac{-2 \pm \sqrt{32}}{2} = -1 \pm 2\sqrt{2}$$

We must verify these against our condition $x \ge 1.5$. The root $-1 - 2\sqrt{2} \approx -3.82$ is rejected, while $-1 + 2\sqrt{2} \approx 1.82$ is accepted.
Thus, our first valid root is $x_1 = 2\sqrt{2} - 1$.

### Case 2: The Left Side of the Boundary

Now, we examine the region where $x < 1.5$. Here, $2x - 3$ is negative, so the modulus outputs the negation of the expression:
$$x^2 - (2x - 3) - 4 = 0$$

Simplifying this yields:
$$x^2 - 2x + 3 - 4 = 0 \Rightarrow x^2 - 2x - 1 = 0$$

Applying the quadratic formula again:
$$x = \frac{2 \pm \sqrt{4 - 4(1)(-1)}}{2} = \frac{2 \pm \sqrt{8}}{2} = 1 \pm \sqrt{2}$$

We check these against the condition $x < 1.5$. The root $1 + \sqrt{2} \approx 2.41$ is rejected, while $1 - \sqrt{2} \approx -0.41$ is accepted.
Thus, our second valid root is $x_2 = 1 - \sqrt{2}$.

### Final Calculation

We have identified the two valid roots: $x_1 = 2\sqrt{2} - 1$ and $x_2 = 1 - \sqrt{2}$. The problem requires the sum of the squares of these roots.

First, we square $x_1$:
$$x_1^2 = (2\sqrt{2} - 1)^2 = 8 - 4\sqrt{2} + 1 = 9 - 4\sqrt{2}$$

Next, we square $x_2$:
$$x_2^2 = (1 - \sqrt{2})^2 = 1 - 2\sqrt{2} + 2 = 3 - 2\sqrt{2}$$

Summing these squares together:
$$(9 - 4\sqrt{2}) + (3 - 2\sqrt{2}) = 12 - 6\sqrt{2}$$

Factoring the result, we arrive at the final answer:
**$6(2 - \sqrt{2})$**

Animated Solution for Mathematics - Quadratic Equations: The sum, of the squares of all the roots of the equation $x^{2} + ∣2 x - 3∣ - 4 = 0$ is

Select Answer:

Visualized Solution

The Modulus Equation

Identifying the Critical Point

Case 1: $x \geq \frac{3}{2}$

Solving Case 1 Quadratic

Validating Case 1 Roots

Case 2: $x < \frac{3}{2}$

Solving Case 2 Quadratic

Validating Case 2 Roots

Setting Up the Sum of Squares

Calculating the Squares

Final Summation

The Sigma Insight: Solution of Quadratic Equations

Analyzing the Setup

Case 1

Case 2

Final Calculation

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Animated Solution for Mathematics - Quadratic Equations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum, of the squares of all the roots of the equation x2+∣2x−3∣−4=0 is