x_1 and x_2 is root log_2 x^((1+^2 log x)) =2, the value is x_1 +x_2 = ... a. 2(1/4) b. 2(1/2) c. 4(1/4) d. 4(1/2) e. 6(1/4)


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Question Number 154989 by bagjagugum123 last updated on 24/Sep/21

$x_{1} a n d x_{2} i s r o o t \log_{2} x^{(1 +^{2} \log x)} = 2, t h e v a l u e$ $i s x_{1} + x_{2} = . . .$ $a . 2 \frac{1}{4}$ $b . 2 \frac{1}{2}$ $c . 4 \frac{1}{4}$ $d . 4 \frac{1}{2}$ $e . 6 \frac{1}{4}$
	Answered by john_santu last updated on 24/Sep/21
	$⇔ (log _2 x+1)log _2 x = 2 let y=log _2 x ⇒y^2 +y−2=0 ⇒(y+2)(y−1)=0 { ((y=−2⇒x_1 =(1/4))),((y=1⇒x_2 =2)) :} ⇒x_1 +x_2 =(9/4)$
	$\Leftrightarrow (\log_{2} x + 1) \log_{2} x = 2$ $l e t y = \log_{2} x$ $\Rightarrow y^{2} + y - 2 = 0$ $\Rightarrow (y + 2) (y - 1) = 0$ ${\begin{cases} y = - 2 \Rightarrow x_{1} = \frac{1}{4} \\ y = 1 \Rightarrow x_{2} = 2 \end{cases} \Rightarrow x_{1} + x_{2} = \frac{9}{4}$
	Commented by bagjagugum123 last updated on 24/Sep/21

	$t h a n k y o u v e r y m u c h S i r$
	Answered by puissant last updated on 24/Sep/21

	$\Rightarrow (1 + {l o g}_{2} x) {l o g}_{2} x = 2$ $t = {l o g}_{2} x \to t^{2} + t - 2 = 0$ $\Rightarrow t_{1} = - 2; t_{2} = 1$ $\frac{{l n x}_{1}}{l n 2} = - 2 \Rightarrow l n x = - 2 l n 2 \Rightarrow x = \frac{1}{4}$ $\frac{{l n x}_{2}}{l n 2} = 1 \Rightarrow {l n x}_{2} = l n 2 \Rightarrow x_{2} = 2$ $∴∵ x_{1} + x_{2} = \frac{1}{4} + \frac{8}{4} = \frac{9}{4} . .$
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