Question

⭐HOLA!!⭐

Solve the following:-

$(1)4 {}^{(2 {x}^{2} + x - 6)} = 1$
and
$(2) ({x}^{2} - 3x + 1) {}^{(2x {}^{2} + x - 6) } = 1$


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Answer 1

$(1)4 {}^{(2 {x}^{2} + x - 6)} = 1$

$4 {}^{(2 {x}^{2} - 3x + 4x - 6)} = 1$

$4 {}^{(x(2x - 3) + 2(2x - 3)} = 1$

$4 {}^{(x+2) (2x - 3)} = 1$

Using differentiation,

$4 {}^{(1) (2)} = 1$

Answer 2

Answer :-

(1) .

$4^{(2x^2 + x - 6)} = 1$

$\implies 4^{(2x^2 + x - 6)} = 4^0$

◾As base are same

$\implies 2x^2 + x - 6 = 0$

◾Now factorization of the Quadratic equation.

$\implies 2x(x+2) - 3(x+2) = 0$

$\implies (2x-3)(x+2) = 0$

⏩Hence ,

x = -2 , $\bold{\dfrac{3}{2}}$

(2).

$(x^2 -3x + 1)^{(2x^2 +x-6)} = 1$

$\implies (x^2 -3x + 1)^{(2x^2 +x-6)} = (x^2 -3x + 1)^{0}$

◾As base are same

$\implies 2x^2 + x - 6 = 0$

◾Now factorization of the Quadratic equation.

$\implies 2x(x+2) - 3(x+2) = 0$

$\implies (2x-3)(x+2) = 0$

⏩Hence ,

x = -2 , $\bold{\dfrac{3}{2}}$