JEE Main 2019 (10 April Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Matrices and Determinants: The sum of the real roots of the equation $x 2 - 3 - 6 - 3 x 2 x - 1 x - 3 x + 2 = 0$ , is equal to :

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark: Visualize the determinant as a function of x. Do not rush into expanding it blindly. Notice the structure of the matrix; can you see any patterns or potential simplifications before you start the heavy algebraic lifting?

Hint 2

Hint 2: The Arsenal: You are dealing with a cubic polynomial. Instead of finding each root individually, recall Vieta's formulas. They relate the coefficients of a polynomial directly to the sum and product of its roots. This is your most powerful tool here.

Hint 3

Hint 3: The Execution: After expanding the determinant, you will arrive at a cubic equation of the form ax^3 + bx^2 + cx + d = 0. Identify the coefficients a, b, c, and d. Remember that the sum of the roots is given by -b/a. Ensure you verify that the roots are indeed real, as the question specifically asks for the sum of the real roots.

Visualized Solution

Analyze the Determinant Equation

Given Equation: $x 2 - 3 - 6 - 3 x 2 x - 1 x - 3 x + 2 = 0$
Objective: Find the sum of real roots of the resulting polynomial.

Choose Expansion Path

Expanding along Row 1 ( $R_{1}$ ).
Sign Convention: $+ - +$
Expansion Formula: $a_{11} M_{11} - a_{12} M_{12} + a_{13} M_{13} = 0$

Set Up the Expansion Terms

Expansion Setup:
$x - 3 x 2 x x - 3 x + 2 - (- 6) 2 - 3 x - 3 x + 2 + (- 1) 2 - 3 - 3 x 2 x = 0$

Expand the First Minor

First Term: $x [(- 3 x) (x + 2) - (2 x) (x - 3)]$
Simplifying inside: $x [- 3 x^{2} - 6 x - (2 x^{2} - 6 x)]$

Simplify the First Term

Combining terms: $x [- 3 x^{2} - 6 x - 2 x^{2} + 6 x]$
Result of First Term: $x [- 5 x^{2}] = - 5 x^{3}$

Expand the Second Minor

Second Term: $+ 6 [2 (x + 2) - (- 3) (x - 3)]$
Simplifying inside: $6 [2 x + 4 - (- 3 x + 9)]$
Further simplifying: $6 [2 x + 4 + 3 x - 9]$

Simplify the Second Term

Combining terms: $6 [5 x - 5]$
Result of Second Term: $30 x - 30$

Expand the Third Minor

Third Term: $- 1 [2 (2 x) - (- 3) (- 3 x)]$
Simplifying inside: $- 1 [4 x - 9 x]$
Result of Third Term: $- 1 [- 5 x] = 5 x$

Form the Cubic Equation

Combining all parts: $- 5 x^{3} + (30 x - 30) + 5 x = 0$
Simplified Equation: $- 5 x^{3} + 35 x - 30 = 0$

Simplify the Equation

Divide by $- 5$ : $\frac{- 5 x ^{3}}{- 5} + \frac{35 x}{- 5} - \frac{30}{- 5} = 0$
Final Polynomial: $x^{3} - 7 x + 6 = 0$
Note: The $x^{2}$ term is missing, so its coefficient is $0$ .

Apply Vieta's Formulas

For $a x^{3} + b x^{2} + c x + d = 0$ , Sum of roots $(α + β + γ) = - \frac{b}{a}$
In our equation $x^{3} + 0 x^{2} - 7 x + 6 = 0$ :
$a = 1$ , $b = 0$ , $c = - 7$ , $d = 6$
Sum of roots $= - \frac{0}{1} = 0$

Verify Real Roots

Check for roots: $f (1) = 1^{3} - 7 (1) + 6 = 0 ⟹ (x - 1)$ is a factor.
Factorizing: $(x - 1) (x^{2} + x - 6) = 0$
Further factorizing: $(x - 1) (x + 3) (x - 2) = 0$
Roots are $x = 1, 2, - 3$ . All are real.

Final Conclusion

Sum of real roots $= 1 + 2 + (- 3) = 0$
Final Answer: 0
Key Takeaway: Missing terms in a polynomial imply a zero coefficient, which directly affects the sum or product of roots via Vieta's formulas.

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The Sigma Insight: Properties of Determinants

Solution Diagram

Analyzing the Setup

To find the sum of the real roots of the equation, we must first evaluate the determinant of the given

3 \times 3

matrix:

x 2 - 3 - 6 - 3 x 2 x - 1 x - 3 x + 2 = 0

We will expand this determinant along the first row using the standard sign convention of plus, minus, plus.

The Expansion Journey

For the first term,

x

, we multiply by its minor:

x - 3 x 2 x x - 3 x + 2 = x [(- 3 x) (x + 2) - (2 x) (x - 3)]

Simplifying the expression inside the brackets:

x [- 3 x^{2} - 6 x - (2 x^{2} - 6 x)] = x [- 5 x^{2}] = - 5 x^{3}

For the second term, we use

- (- 6) = + 6

multiplied by its minor:

6 2 - 3 x - 3 x + 2 = 6 [2 (x + 2) - (- 3) (x - 3)]

This simplifies to:

6 [2 x + 4 + 3 x - 9] = 6 [5 x - 5] = 30 x - 30

For the third term, we take

- 1

multiplied by its minor:

- 1 2 - 3 - 3 x 2 x = - 1 [2 (2 x) - (- 3) (- 3 x)] = - 1 [4 x - 9 x] = 5 x

The Cubic Emergence

Combining these components, we obtain the following equation:

- 5 x^{3} + (30 x - 30) + 5 x = 0

Grouping like terms yields:

- 5 x^{3} + 35 x - 30 = 0

Dividing the entire equation by

- 5

results in the monic cubic polynomial:

x^{3} - 7 x + 6 = 0

Note that the coefficient of the

x^{2}

term is

0

The Vieta's Insight

According to Vieta's formulas, for any cubic equation of the form

a x^{3} + b x^{2} + c x + d = 0

, the sum of the roots is given by

- \frac{b}{a}

In our equation,

a = 1

and

b = 0

. Therefore, the sum of the roots is:

- \frac{0}{1} = 0

The Reality Check

To ensure all roots are real, we test

x = 1

and find

1^{3} - 7 (1) + 6 = 0

, confirming that

(x - 1)

is a factor. Dividing the cubic by

(x - 1)

gives:

(x - 1) (x^{2} + x - 6) = 0

Factoring the quadratic further, we get:

(x - 1) (x + 3) (x - 2) = 0

The roots are

1, 2, - 3

. Since all roots are real, their sum is

1 + 2 - 3 = 0

The sum of the real roots is 0.

Common Traps & FAQs

Wait, why do we reject the negative root here?

Actually, we do not reject any roots! The question asks for the sum of the *real* roots. In this specific problem, all three roots of the cubic equation are real. If we had encountered complex roots, we would have had to exclude them from the sum. Always check the nature of the roots using the discriminant or by testing factors.

Why can't we just use energy conservation?

That is a great question, but energy conservation is a principle used in physics problems involving motion or forces. This is a pure mathematics problem involving determinants and polynomial roots. Here, we use the properties of determinants and Vieta's formulas to find the solution.

Is there a faster way to find the sum of the roots without expanding the whole determinant?

While you could potentially use properties of determinants to simplify the matrix before expansion, expanding it is quite straightforward here. Once you have the cubic equation, Vieta's formulas provide the fastest route to the sum of the roots without needing to solve for each root individually. However, verifying the roots are real is a crucial step you should never skip!

Animated Solution for Mathematics - Matrices and Determinants: The sum of the real roots of the equation $x 2 - 3 - 6 - 3 x 2 x - 1 x - 3 x + 2 = 0$ , is equal to :

Select Answer:

Visualized Solution

Analyze the Determinant Equation

Choose Expansion Path

Set Up the Expansion Terms

Expand the First Minor

Simplify the First Term

Expand the Second Minor

Simplify the Second Term

Expand the Third Minor

Form the Cubic Equation

Simplify the Equation

Apply Vieta's Formulas

Verify Real Roots

Final Conclusion

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Expansion Journey

The Cubic Emergence

The Vieta's Insight

The Reality Check

Similar Questions

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Given that $x = - 9$ is a root of $x 27 3 x 6 72 x = 0$ the other two roots are ......... and .........

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Animated Solution for Mathematics - Matrices and Determinants: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum of the real roots of the equation ​x2−3​−6−3x2x​−1x−3x+2​​=0, is equal to :

Select Answer:

Visualized Solution

Analyze the Determinant Equation

Choose Expansion Path

Set Up the Expansion Terms

Expand the First Minor

Simplify the First Term

Expand the Second Minor

Simplify the Second Term

Expand the Third Minor

Form the Cubic Equation

Simplify the Equation

Apply Vieta's Formulas

Verify Real Roots

Final Conclusion

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Expansion Journey

The Cubic Emergence

The Vieta's Insight

The Reality Check

Similar Questions

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Animated Solution for Mathematics - Matrices and Determinants: The sum of the real roots of the equation $x 2 - 3 - 6 - 3 x 2 x - 1 x - 3 x + 2 = 0$ , is equal to :

The number of distinct real roots of $sin x cos x cos x cos x sin x cos x cos x cos x sin x = 0$ in the interval $- π /4 \leq x \leq π /4$ is

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The number of distinct real roots of $sin x cos x cos x cos x sin x cos x cos x cos x sin x = 0$ in the interval $- \frac{π}{4} \leq x \leq \frac{π}{4}$ is:

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If $Δ = x - 2 2 x - 3 3 x - 5 2 x - 3 3 x - 4 5 x - 8 3 x - 4 4 x - 5 10 x - 17 = A x^{3} + B x^{2} + C x + D$ , then $B + C$ is equal to :

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