The sum of all the real solutions of the equation (x+3)(6x²+28x+30)=5-2 (6x+10)(x²+6x+9)$\log_{(x+3)}(6x^{2}+28x+30)=5-2\log_{(6x+10)}(x^{2}+6x+9)$ is equal to:

Related Formula

Modulus definition rule:

Core Logic

Analyze the roots by splitting into cases around the critical threshold x = (3)/(2)$x = \frac{3}{2}$:

Case I: x ≥ (3)/(2)$x \ge \frac{3}{2}$

(We select the positive root since 2√(2)-1 ≥ 1.5$2\sqrt{2}-1 \ge 1.5$).

Step 1: Evaluating the alternate domain branch

Case II: x < (3)/(2)$x < \frac{3}{2}$

(We select 1-√(2)$1-\sqrt{2}$ since it satisfies the inequality constraint).

Step 2: Summing the Squares of the Roots

Pattern Recognition

Always validate absolute root values against their domain restrictions to avoid including phantom solutions.

Chapter Mix

Class 11 Maths: Quadratic Equations

Related Formula

Newton's Theorem for roots: If α, β$\alpha, \beta$ satisfy ax² + bx + c = 0$ax^2 + bx + c = 0$ and Sₙ = αⁿ + βⁿ$S_n = \alpha^n + \beta^n$, then:

Core Logic

Apply Newton's Theorem directly to the first equation x² + √(3)x - 16 = 0$x^2 + \sqrt{3}x - 16 = 0$:

For n = 25$n = 25$:

Dividing both sides by 2P₂₃$2P_{23}$:

Step 1: Evaluation of the Second Part

For the second equation x² + 3x - 1 = 0$x^2 + 3x - 1 = 0$:

For n = 25$n = 25$:

Dividing both sides by Q₂₄$Q_{24}$:

Step 2: Total Calculation

Add both evaluated components:

Pattern Recognition

Shortcut: High sequential indices (25, 24, 23$25, 24, 23$) indicate recurrence via Newton's Theorem. Relate Pₙ$P_n$ and Qₙ$Q_n$ directly to their characteristic quadratic polynomials to evaluate the ratios in one step without calculating powers.

Evaluation Rubric / Model Answer

Option (C)

Chapter Mix

Class 11 Mathematics: Quadratic Equations

Related Formula

For quadratic equations with integer coefficients to have integral roots, the discriminant D = b² - 4ac$D = b^2 - 4ac$ must be a perfect square.

Core Logic

Rewrite using perfect square completing methods:

For x$x$ to be an integer, n + 4$n + 4$ must be a perfect square. Given range constraint 20 ≤ n ≤ 100$20 \le n \le 100$:

Step 1: Identify Perfect Squares in Range

Find perfect squares between 24 and 104: 5² = 25$5^2 = 25$ 6² = 36$6^2 = 36$ 7² = 49$7^2 = 49$ 8² = 64$8^2 = 64$ 9² = 81$9^2 = 81$ 10² = 100$10^2 = 100$

This gives exactly 6 distinct valid perfect squares.

Step 2: Conclusion

Thus, there are exactly 6 distinct integer values for n$n$.

Pattern Recognition

Completing the square provides intuitive bounds quicker than running full discriminant inequalities. Match integer root sets directly to explicit numerical sequence counts.

Chapter Mix

Class 10 Mathematics: Quadratic Equations Class 11 Mathematics: Complex Numbers and Quadratic Equations

Related Formula

The definition of modulus function handles sub-intervals via critical points:

Core Logic

The critical points are x = 2$x = 2$ and x = 3$x = 3$. We check the three distinct structural intervals:

Case I: x < 2$x < 2$

Checking domain constraint x < 2$x < 2$: -1 - √(41)2 ≈ (-1 - 6.4)/(2) = -3.7 < 2$\frac{-1 - \sqrt{41}}{2} \approx \frac{-1 - 6.4}{2} = -3.7 < 2$ (Valid root) -1 + √(41)2 ≈ (-1 + 6.4)/(2) = 2.7 < 2$\frac{-1 + \sqrt{41}}{2} \approx \frac{-1 + 6.4}{2} = 2.7 \not< 2$ (Rejected)

Step 1: Intermediate Interval Check

Case II: 2 ≤ x < 3$2 \le x < 3$

Discriminant check: D = (-5)² - 4(1)(10) = 25 - 40 = -15 < 0$D = (-5)^2 - 4(1)(10) = 25 - 40 = -15 < 0$. No real roots exist in this interval.

Step 2: Upper Interval Check

Case III: x ≥ 3$x \ge 3$

Checking domain constraint x ≥ 3$x \ge 3$: -1 + √(33)2 ≈ (-1 + 5.74)/(2) = 2.37 < 3$\frac{-1 + \sqrt{33}}{2} \approx \frac{-1 + 5.74}{2} = 2.37 < 3$ (Rejected) -1 - √(33)2 < 0$\frac{-1 - \sqrt{33}}{2} < 0$ (Rejected)

Thus, only 1 valid real root satisfies the conditional layout across all ranges.

Pattern Recognition

Always perform case-by-case boundaries checks on algebraic roots found inside absolute modulus problems to discard ghost solutions quickly.

Chapter Mix

Class 11 Mathematics: Quadratic Equations

Related Formula

For equations with repeating polynomial expressions, applying a variable substitution like t = P(x)$t = P(x)$ reduces high-degree polynomials down to standard quadratics.

Core Logic

Expand the linear binomial product in the given equation equation:

Rewrite the full expression in terms of the common variable pattern x² - 9x$x^2 - 9x$:

Let t = x² - 9x$t = x^2 - 9x$. Substituting this into the equation gives:

Step 1: Solve the Polynomial for Variable t

Factorize the quadratic expression:

Step 2: Back-substitute and Isolate Rational Roots

Case 1: x² - 9x = -7 x² - 9x + 7 = 0$x^2 - 9x = -7 \implies x^2 - 9x + 7 = 0$ Check the discriminant value: D = (-9)² - 4(1)(7) = 81 - 28 = 53$D = (-9)^2 - 4(1)(7) = 81 - 28 = 53$ (not a perfect square, so the roots are irrational).

Case 2: x² - 9x = -14 x² - 9x + 14 = 0$x^2 - 9x = -14 \implies x^2 - 9x + 14 = 0$ Factorize the quadratic expression:

Both values are rational numbers.

Step 3: Calculate the Product of Rational Roots

Multiply the true rational roots together:

Pattern Recognition

Always check the discriminant D = b² - 4ac$D = b^2 - 4ac$ to filter out irrational radical components whenever the problem specifically asks for the product of rational roots only.

Chapter Mix

Class 11 Mathematics: Quadratic Equations