Solve for x : (5 + 2√6)^(x²-3) + (5 -...

MathematicsTransformation of EquationsJEE Advanced 1985Easy

Solve for $x : (5 + 26)^{x^{2} - 3} + (5 - 26)^{x^{2} - 3} = 10$

Answer:

\pm 2, \pm 2

Visualized Solution (Hindi)

Analyze the Equation

Given equation: $(5 + 26)^{x^{2} - 3} + (5 - 26)^{x^{2} - 3} = 10$
Objective: Solve for the variable $x$ .

Identify Reciprocal Bases

Observe the product of the bases:
$(5 + 26) (5 - 26) = 5^{2} - (26)^{2}$
$= 25 - 24 = 1$
Conclusion: $(5 - 26) = \frac{1}{5 + 2 6}$

Substitution Strategy

Let $t = (5 + 26)^{x^{2} - 3}$
Then $(5 - 26)^{x^{2} - 3} = (\frac{1}{5 + 2 6})^{x^{2} - 3} = \frac{1}{t}$

Form the Quadratic Equation

Substitute into the original equation:
$t + \frac{1}{t} = 10$
Multiply by $t$ : $t^{2} + 1 = 10 t$
Standard form: $t^{2} - 10 t + 1 = 0$

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t

Using quadratic formula: $t = \frac{- b \pm b ^{2} - 4 a c}{2 a}$
$t = \frac{10 \pm ( - 10 ) ^{2} - 4 ( 1 ) ( 1 )}{2 ( 1 )}$
$t = \frac{10 \pm 96}{2} = \frac{10 \pm 4 6}{2}$
$t = 5 \pm 26$

Case 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t = 5 + 26

Set $t = 5 + 26$ :
$(5 + 26)^{x^{2} - 3} = (5 + 26)^{1}$
Comparing exponents: $x^{2} - 3 = 1$

Solve Case 1 for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Transpose $- 3$ : $x^{2} = 1 + 3$
$x^{2} = 4$
Taking square root: $x = \pm 2$

Case 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t = 5 - 26

Set $t = 5 - 26$ :
$(5 + 26)^{x^{2} - 3} = 5 - 26$
Since $5 - 26 = (5 + 26)^{- 1}$ :
$(5 + 26)^{x^{2} - 3} = (5 + 26)^{- 1}$
Comparing exponents: $x^{2} - 3 = - 1$

Solve Case 2 for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Transpose $- 3$ : $x^{2} = - 1 + 3$
$x^{2} = 2$
Taking square root: $x = \pm 2$

Final Solutions

The solutions are $x = \pm 2$ and $x = \pm 2$ .
Key Takeaway: Recognize reciprocal bases $(a + b) (a - b) = 1$ to simplify exponential equations into quadratics.
Next Challenge: Try solving if the equation was equal to $10.1$ instead of $10$ .

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Answer:

x \in [1 - 5, 1) \cup (2, 1 + 5]

Solve for $x : (5 + 26)^{x^{2} - 3} + (5 - 26)^{x^{2} - 3} = 10$

Visualized Solution (Hindi)

Analyze the Equation

Identify Reciprocal Bases

Substitution Strategy

Form the Quadratic Equation

Solve for $t$

Case 1: $t = 5 + 26$

Solve Case 1 for $x$

Case 2: $t = 5 - 26$

Solve Case 2 for $x$

Final Solutions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solve for x:(5+26​)x2−3+(5−26​)x2−3=10

Visualized Solution (Hindi)

Analyze the Equation

Identify Reciprocal Bases

Substitution Strategy

Form the Quadratic Equation

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t

Case 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t=5+26​

Solve Case 1 for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Case 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t=5−26​

Solve Case 2 for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Final Solutions

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Solve for $x : (5 + 26)^{x^{2} - 3} + (5 - 26)^{x^{2} - 3} = 10$

Solve for $t$

Case 1: $t = 5 + 26$

Solve Case 1 for $x$

Case 2: $t = 5 - 26$

Solve Case 2 for $x$

Solve $∣ x^{2} + 4 x + 3∣ + ∣2 x + 5∣ = 0$

Solve for $x : x + 1 - x - 1 = 1$ .

For $a \leq 0$ , determine all real roots of the equation $x^{2} - 2 a ∣ x - a ∣ - 3 a^{2} = 0$

Solve for $x : 4^{x} - 3^{x - 1/2} = 3^{x + 1/2} - 2^{2 x - 1}$

The sum of all real values of $x$ satisfying the equation $(x^{2} - 5 x + 5)^{x^{2} + 4 x - 60} = 1$ is :

If $α \neq = β$ but $α^{2} = 5 α - 3$ and $β^{2} = 5 β - 3$ then the equation having $α / β$ and $β / α$ as its roots is

Solve for $x$ the following equation : $lo g_{(2 x + 3)} (6 x^{2} + 23 x + 21) = 4 - lo g_{(3 x + 7)} (4 x^{2} + 12 x + 9)$ .

Solve the following equation for $x : 2 lo g_{x} a + lo g_{a x} a + 3 lo g_{a^{2} x} a = 0, a > 0, a \neq = 0$

Find all real values of $x$ which satisfy $x^{2} - 3 x + 2 > 0$ and $x^{2} - 2 x - 4 \leq 0$

Solve for $x : (5 + 26)^{x^{2} - 3} + (5 - 26)^{x^{2} - 3} = 10$

Solve $∣ x^{2} + 4 x + 3∣ + ∣2 x + 5∣ = 0$

Solve for $x : x + 1 - x - 1 = 1$ .

For $a \leq 0$ , determine all real roots of the equation $x^{2} - 2 a ∣ x - a ∣ - 3 a^{2} = 0$

Solve for $x : 4^{x} - 3^{x - 1/2} = 3^{x + 1/2} - 2^{2 x - 1}$

The sum of all real values of $x$ satisfying the equation $(x^{2} - 5 x + 5)^{x^{2} + 4 x - 60} = 1$ is :

If $α \neq = β$ but $α^{2} = 5 α - 3$ and $β^{2} = 5 β - 3$ then the equation having $α / β$ and $β / α$ as its roots is

Solve for $x$ the following equation : $lo g_{(2 x + 3)} (6 x^{2} + 23 x + 21) = 4 - lo g_{(3 x + 7)} (4 x^{2} + 12 x + 9)$ .

Solve the following equation for $x : 2 lo g_{x} a + lo g_{a x} a + 3 lo g_{a^{2} x} a = 0, a > 0, a \neq = 0$

Find all real values of $x$ which satisfy $x^{2} - 3 x + 2 > 0$ and $x^{2} - 2 x - 4 \leq 0$