Core Logic
Restructure the equation to form a quadratic in |√(x) - 3|$|\sqrt{x} - 3|$.
Notice that (x - 6√(x) + 9) = (√(x) - 3)² = |√(x) - 3|²$(x - 6\sqrt{x} + 9) = (\sqrt{x} - 3)^2 = |\sqrt{x} - 3|^2$.
Rewrite the given equation:
(x - 6√(x) + 9) - (2 - √(3))|√(x) - 3| - 2√(3) = 0$$(x - 6\sqrt{x} + 9) - (2 - \sqrt{3})|\sqrt{x} - 3| - 2\sqrt{3} = 0$$
|√(x) - 3|² - (2 - √(3))|√(x) - 3| - 2√(3) = 0$$|\sqrt{x} - 3|^2 - (2 - \sqrt{3})|\sqrt{x} - 3| - 2\sqrt{3} = 0$$
Step 1: Solve the Quadratic
Let u = |√(x) - 3|$u = |\sqrt{x} - 3|$. The equation is u² - (2 - √(3))u - 2√(3) = 0$u^2 - (2 - \sqrt{3})u - 2\sqrt{3} = 0$.
Factorizing gives:
(u - 2)(u + √(3)) = 0$$(u - 2)(u + \sqrt{3}) = 0$$
So, u = 2$u = 2$ or u = -√(3)$u = -\sqrt{3}$.
Since u = |√(x) - 3|$u = |\sqrt{x} - 3|$ cannot be negative, we reject u = -√(3)$u = -\sqrt{3}$.
Thus, |√(x) - 3| = 2$|\sqrt{x} - 3| = 2$.
Step 2: Find x (Roots)
Solve |√(x) - 3| = 2$|\sqrt{x} - 3| = 2$:
√(x) - 3 = 2 or √(x) - 3 = -2$$\sqrt{x} - 3 = 2 \quad \text{or} \quad \sqrt{x} - 3 = -2$$
√(x) = 5 or √(x) = 1$$\sqrt{x} = 5 \quad \text{or} \quad \sqrt{x} = 1$$
Squaring gives x = 25$x = 25$ or x = 1$x = 1$.
Given α < β$\alpha < \beta$, we have α = 1$\alpha = 1$ and β = 25$\beta = 25$.
Step 3: Evaluate Target Expression
Now compute √((β)/(α)) + √(αβ)$\sqrt{\frac{\beta}{\alpha}} + \sqrt{\alpha\beta}$:
= √((25)/(1)) + √(1 · 25) = 5 + 5 = 10$$= \sqrt{\frac{25}{1}} + \sqrt{1 \cdot 25} = 5 + 5 = 10$$
Pattern Recognition
Grouping algebraic terms (like x - 6√(x)$x - 6\sqrt{x}$) and a lone constant (+9$+9$) to form perfect squares is a hallmark of radical equations disguised as quadratics.
Chapter Mix
Class 11 Maths: Quadratic Equations
