If α$\alpha$ and β$\beta$ (α < β$\alpha < \beta$) are the roots of the equation (- 2 + √(3)) (| √(x) - 3 |) + (x - 6 √(x)) + (9 - 2 √(3)) = 0$(- 2 + \sqrt {3}) (| \sqrt {x} - 3 |) + (x - 6 \sqrt {x}) + (9 - 2 \sqrt {3}) = 0$, x ≥ 0$x \geq 0$, then √((β)/(α)) + √(αβ)$\sqrt{\frac{\beta}{\alpha}} + \sqrt{\alpha\beta}$ is equal to:

Core Logic

Let the time taken by mason A alone to complete the work be x$x$ days. Mason B takes x + 24$x + 24$ days. Work done by A in 1 day = (1)/(x)$\frac{1}{x}$ Work done by B in 1 day = (1)/(x + 24)$\frac{1}{x + 24}$

Step 1: Set up the Rate Equation

Since together they finish in 22.5$22.5$ days, their combined work in 1$1$ day is (1)/(22.5) = (1)/(45/2) = (2)/(45)$\frac{1}{22.5} = \frac{1}{45/2} = \frac{2}{45}$. So, (1)/(x) + (1)/(x + 24) = (2)/(45)$\frac{1}{x} + \frac{1}{x + 24} = \frac{2}{45}$

Step 2: Solve the Quadratic

Multiply to clear denominators:

Factorizing:

Since time cannot be negative, we reject x = -15$x = -15$. So, x = 36$x = 36$ days.

Pattern Recognition

Standard "Time & Work" reciprocal addition resolves purely to a clean factorizable quadratic. Setting the faster worker to x$x$ prevents dealing with negative bounds in factors.

Chapter Mix

Class 11 Maths: Basic Mathematics

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Core Logic

Factor the arguments in the logarithmic equation: 6x² + 28x + 30 = (x+3)(6x+10)$6x^2 + 28x + 30 = (x+3)(6x+10)$ x² + 6x + 9 = (x+3)²$x^2 + 6x + 9 = (x+3)^2$

Substitute these into the equation:

Step 1: Variable Substitution

Let A = (x+3)(6x+10)$A = \log_{(x+3)}(6x+10)$. The equation transforms to:

Substitute A = 2$A = 2$ back:

Step 2: Checking Domain Validity

For x = 1$x = 1$: Base x+3 = 4 > 0, ≠ 1$x+3 = 4 > 0, \neq 1$. Base 6x+10 = 16 > 0, ≠ 1$6x+10 = 16 > 0, \neq 1$. Valid solution.

For x = -1$x = -1$: Base x+3 = 2 > 0, ≠ 1$x+3 = 2 > 0, \neq 1$. Base 6x+10 = 4 > 0, ≠ 1$6x+10 = 4 > 0, \neq 1$. Valid solution.

Sum of all real roots = 1 + (-1) = 0$= 1 + (-1) = 0$.

Pattern Recognition

When dealing with logarithms containing polynomial bases and arguments, always check if they are directly factorable into each other. A substitution like A + B/A = C$A + B/A = C$ will often emerge.

Chapter Mix

Class 11 Maths: Quadratic Equations Class 11 Maths: Functions

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Core Logic

Step 1: Case 1 (x > 1)

For x > 1$x > 1$: x(x+3) + (x-1) - 2 = 0$x(x+3) + (x-1) - 2 = 0$ x² + 3x + x - 3 = 0 ⇒ x² + 4x - 3 = 0$x^2 + 3x + x - 3 = 0 \Rightarrow x^2 + 4x - 3 = 0$ x = -4 ± √(16 + 12)2 = -2 ± √(7)$x = \frac{-4 \pm \sqrt{16 + 12}}{2} = -2 \pm \sqrt{7}$ Since √(7) ≈ 2.64$\sqrt{7} \approx 2.64$, x = -2 + 2.64 = 0.64$x = -2 + 2.64 = 0.64$, which is not > 1$> 1$. Both rejected.

Step 2: Case 2 (-3 <= x <= 1)

For -3 ≤ x ≤ 1$-3 \le x \le 1$: x(x+3) - (x-1) - 2 = 0$x(x+3) - (x-1) - 2 = 0$ x² + 3x - x + 1 - 2 = 0 ⇒ x² + 2x - 1 = 0$x^2 + 3x - x + 1 - 2 = 0 \Rightarrow x^2 + 2x - 1 = 0$ x = -2 ± √(4 + 4)2 = -1 ± √(2)$x = \frac{-2 \pm \sqrt{4 + 4}}{2} = -1 \pm \sqrt{2}$ √(2) ≈ 1.41$\sqrt{2} \approx 1.41$. x = -1 + 1.41 = 0.41$x = -1 + 1.41 = 0.41$ (Accepted) x = -1 - 1.41 = -2.41$x = -1 - 1.41 = -2.41$ (Accepted) (2 solutions)

Step 3: Case 3 (x < -3)

For x < -3$x < -3$: x(-x-3) - (x-1) - 2 = 0$x(-x-3) - (x-1) - 2 = 0$ -x² - 3x - x + 1 - 2 = 0 ⇒ -x² - 4x - 1 = 0 ⇒ x² + 4x + 1 = 0$-x^2 - 3x - x + 1 - 2 = 0 \Rightarrow -x^2 - 4x - 1 = 0 \Rightarrow x^2 + 4x + 1 = 0$ x = -4 ± √(16 - 4)2 = -2 ± √(3)$x = \frac{-4 \pm \sqrt{16 - 4}}{2} = -2 \pm \sqrt{3}$ x = -2 + 1.732 = -0.268$x = -2 + 1.732 = -0.268$ (Rejected, not < -3$< -3$) x = -2 - 1.732 = -3.732$x = -2 - 1.732 = -3.732$ (Accepted, < -3$< -3$) (1 solution) Total valid real solutions = 2 + 1 = 3$2 + 1 = 3$.

Pattern Recognition

Modulus equations involving polynomials are best resolved by strictly zoning the number line via critical points, verifying root validity against the respective zone boundaries.

Chapter Mix

Class 11 Maths: Complex Numbers and Quadratic Equations

Related Formula

Core Logic

Substitute x² = t$x^2 = t$. The equation becomes a quadratic in t$t$:

For the original quartic equation x⁴ - ax² + 9 = 0$x^4 - ax^2 + 9 = 0$ to have 4 real and distinct roots, the quadratic equation in t$t$ must have 2 distinct positive real roots (since x = ± √(t)$x = \pm \sqrt{t}$ requires t > 0$t > 0$).

Step 1: Discriminant Condition

Condition 1: Roots must be real and distinct (D > 0$D > 0$)

a² - 36 > 0$a^2 - 36 > 0$

Step 2: Location of Roots Condition

Condition 2: Sum of roots must be positive (since both roots are positive)

Condition 3: Product of roots must be positive

Step 3: Intersection and Conclusion

Taking the intersection of all conditions: a in (-∞, -6) (6, ∞)$a \in (-\infty, -6) \cup (6, \infty)$ AND a > 0$a > 0$.

Intersection yields: a in (6, ∞)$a \in (6, \infty)$.

The smallest positive integral value in this interval is 7$7$.

Pattern Recognition

Bi-quadratic equations x⁴ + Bx² + C = 0$x^4 + Bx^2 + C = 0$ map cleanly to t² + Bt + C = 0$t^2 + Bt + C = 0$. The nature of x$x$ roots depends entirely on the signs of t$t$ roots. 4 real distinct x$x$ roots ≡$\equiv$ 2 positive distinct t$t$ roots.

Chapter Mix

Class 11 Maths: Quadratic Equations