7-4-3-x-2-4x-3-7-4-3-x-2-4x-3-14-

Question Number 9827 by tawakalitu last updated on 06/Jan/17

{(7 - 4 \sqrt{3})}^{x^{2} - 4 x + 3} + {(7 + 4 \sqrt{3})}^{x^{2} - 4 x + 3} = 14

Answered by ridwan balatif last updated on 06/Jan/17

7 - 4 \sqrt{3} = (7 - 4 \sqrt{3}) \frac{(7 + 4 \sqrt{3})}{(7 + 4 \sqrt{3})}

7 - 4 \sqrt{3} = \frac{49 - 48}{7 + 4 \sqrt{3}}

7 - 4 \sqrt{3} = \frac{1}{7 + 4 \sqrt{3}}

let : {(7 - 4 \sqrt{3})}^{x^{2} - 4 x + 3} = a

a + \frac{1}{a} = 14

a^{2} + 1 = 14 a

a^{2} - 14 a + 1 = 0

D = B^{2} - 4 AC

= {(- 14)}^{2} - 4 \times 1 \times 1

= 192

a_{1, 2} = \frac{- b \pm \sqrt{D}}{2 a}

= \frac{14 \pm 8 \sqrt{3}}{2}

= 7 \pm 4 \sqrt{3}

a_{1} = 7 + 4 \sqrt{3} and a_{2} = 7 - 4 \sqrt{3}

first (a_{1} = 7 - 4 \sqrt{3})

{(7 - 4 \sqrt{3})}^{x^{2} - 4 x + 3} = {(7 - 4 \sqrt{3})}^{1}

x^{2} - 4 x + 3 = 1

x^{2} - 4 x + 2 = 0

D = b^{2} - 4 ac

= 16 - 4 \times 1 \times 2

= 8

x_{1, 2} = \frac{4 \pm 2 \sqrt{2}}{2}

= 2 \pm \sqrt{2}

x_{1} = 2 + \sqrt{2} and x_{2} = 2 - \sqrt{2}

continued

Commented by sandy_suhendra last updated on 06/Jan/17

{(7 - 4 \sqrt{3})}^{x^{2} - 4 x + 3} = 7 + 4 \sqrt{3}

{(7 - 4 \sqrt{3})}^{x^{2} - 4 x + 3} = {(7 - 4 \sqrt{3})}^{- 1}

x^{2} - 4 x + 3 = - 1

x^{2} - 4 x + 4 = 0

{(x - 2)}^{2} = 0

x = 2

Commented by tawakalitu last updated on 06/Jan/17

wow, i really appreciate . God bless you .