Related Formula
Conjugate Surd Identity:
(√(a) + √(b))(√(a) - √(b)) = a - b$$(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = a - b$$
Core Logic
Observe the base components of the exponents:
(√(3) + √(2))(√(3) - √(2)) = 3 - 2 = 1$$(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2}) = 3 - 2 = 1$$
Therefore, we can express one base as the reciprocal of the other:
√(3) - √(2) = 1√(3) + √(2)$$\sqrt{3} - \sqrt{2} = \frac{1}{\sqrt{3} + \sqrt{2}}$$
The equation becomes:
(√(3) + √(2))x + 1(√(3) + √(2))x = 10$$(\sqrt{3} + \sqrt{2})^{x} + \frac{1}{(\sqrt{3} + \sqrt{2})^{x}} = 10 $$
Step 1: Formulate the Quadratic Equation
Let (√(3) + √(2))x = t$(\sqrt{3} + \sqrt{2})^{x} = t$. Then:
t + (1)/(t) = 10 t² - 10t + 1 = 0$$t + \frac{1}{t} = 10 \implies t^{2} - 10t + 1 = 0 $$
Solving for t$t$ using the quadratic formula:
t = 10 ± (-10)² - 4(1)(1)2 = 10 ± √(96)2 = 5 ± 2√(6)$$t = \frac{10 \pm \sqrt{(-10)^{2} - 4(1)(1)}}{2} = \frac{10 \pm \sqrt{96}}{2} = 5 \pm 2\sqrt{6} $$
Step 2: Solve for x
Notice that (5 ± 2√(6))$(5 \pm 2\sqrt{6})$ can be written as square powers of the original base:
(√(3) ± √(2))² = 3 + 2 ± 2√(6) = 5 ± 2√(6)$$(\sqrt{3} \pm \sqrt{2})^{2} = 3 + 2 \pm 2\sqrt{6} = 5 \pm 2\sqrt{6}$$
Thus, we have:
- For t = 5 + 2√(6) (√(3) + √(2))x = (√(3) + √(2))² x = 2$t = 5 + 2\sqrt{6} \implies (\sqrt{3} + \sqrt{2})^{x} = (\sqrt{3} + \sqrt{2})^{2} \implies x = 2$
- For t = 5 - 2√(6) (√(3) + √(2))x = (√(3) - √(2))² = (√(3) + √(2))⁻² x = -2$t = 5 - 2\sqrt{6} \implies (\sqrt{3} + \sqrt{2})^{x} = (\sqrt{3} - \sqrt{2})^{2} = (\sqrt{3} + \sqrt{2})^{-2} \implies x = -2$
Therefore, the distinct real solutions are x = 2$x = 2$ and x = -2$x = -2$.
The number of elements in set S$S$ is 2$2$.
Pattern Recognition
Sees: Rational conjugate bases added with inverse matching variables.
Shortcut: Whenever you see an equation of the form A^x + B^x = C$A^x + B^x = C$ where AB = 1$AB = 1$, the solution will always be symmetric (± x₀$\pm x_0$). Checking x=2$x=2$ explicitly gives (√(3)+√(2))² + (√(3)-√(2))² = (5+2√(6)) + (5-2√(6)) = 10$(\sqrt{3}+\sqrt{2})^2 + (\sqrt{3}-\sqrt{2})^2 = (5+2\sqrt{6}) + (5-2\sqrt{6}) = 10$, confirming ± 2$\pm 2$ immediately.
Chapter Mix
Class 11 Mathematics: Quadratic Equations
Class 9 Mathematics: Number Systems (Rationalization)
