Question

Sa se rezolve ecuatia:
(x²-16x)²-2(x²-16x)-63=0

Answer 1

(x²-16x)²-2(x²-16x)-63=0

Notam x²-16x=t

=> t² - 2t - 63 = 0
Δ=4+252 = 256

=> $t_{1,2} = \frac{2+- \sqrt{256}}{2} = \frac{2+-16}{2}$
=> t = -7,   t = 9

(1) t = -7 => x²-16x = -7 => x²-16x + 7 = 0
Δ= 256 - 28 = 228

=> 
$x_{1,2} = \frac{16+- \sqrt{228} }{2} \\ x_{1} = \frac{16-2 \sqrt{57}}{2}= 8 + \sqrt{57} (I) \\ x_{2} = 8 - \sqrt{57} (II)$

(2) t = 9 => x²-16x=9 => x²-16x-9=0
Δ=256 + 36 = 292

$x_{1,2} = \frac{16+- \sqrt{292} }{2} \\ x_{1} = \frac{16-2 \sqrt{73}}{2}= 8 + \sqrt{73} (III) \\ x_{2} = 8 - \sqrt{73} (IV)$

Din (I),(II),(III),(IV) => x∈{8+-√57; 8+-√73}

Answer 2

$\displaystyle Notam~t=x^2-16x,~iar~ecuatia~devine: \\ \\ t^2-2t-63=0 \Leftrightarrow (t+7)(t-9)=0 \Rightarrow t \in \{-7,9\}. \\ \\ Deci~x^2-16x \in \{-7,9 \}. \\ \\ i)~x^2-16x=-7 \Leftrightarrow x^2-16x+7=0. \\ \\ \Delta=(-16)^2-4 \cdot 1 \cdot 7=228=4 \cdot 57. \\ \\ x_{1,2}= \frac{16 \pm \sqrt{4 \cdot 57}}{2}= \frac{16 \pm 2 \sqrt{57}}{2}=8 \pm \sqrt{57}. \\ \\ ii)~x^2-16x=9 \Leftrightarrow x^2-16x-9=0. \\ \\ \Delta'=(-16)^2-4 \cdot 1 \cdot (-9)=292=4 \cdot 73.$

$\displaystyle x_{3,4}= \frac{16 \pm \sqrt{4 \cdot 73}}{2}= \frac{16 \pm 2 \sqrt{73}}{2}=8 \pm \sqrt{73}. \\ \\ Solutie:~ x \in \left \{ 8 \pm \sqrt{57}~;~8 \pm \sqrt{73} \right \}.$