Find the sum of all the real number x that satisfy (2x^2 +5x+1)^(2x−3) =1


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Question Number 144260 by ZiYangLee last updated on 23/Jun/21

$Find the sum of all the real number$ $x that satisfy {(2 x^{2} + 5 x + 1)}^{2 x - 3} = 1$
	Answered by Ar Brandon last updated on 24/Jun/21

	$2 x - 3 = 0 \lor {2 x}^{2} + 5 x + 1 = 1$ $x = \frac{3}{2} \lor x = 0, x = - \frac{5}{2}$ ${2 x}^{2} + 5 x + 1 > 0$
	Answered by ArielVyny last updated on 24/Jun/21
	$solution 1 (1)^α =1 2x^2 +5x+1=1 x(2x+5)=0 x=0 x=−(5/2) solution 2 α^0 =1 with α#0 2x−3=0 x=(3/2) solution 3 (−1)^(2n) =1 2x^2 +5x+1=−1 2x^2 +5x+2=0 2[(x+(5/2))^2 −((25)/4)+(8/4)]=0 2[(x+(5/2))^2 −((17)/4)]=0 2[(x+(5/2)−((√(17))/2))(x+(5/2)+((√(17))/2))]=0 x_1 =−(5/2)+((√(17))/2) and x_2 =−(5/2)−((√(17))/2) then x_1 and x_2 are not peer so in case 3 there is not solution S={0.(3/2).−(5/2)} we have 0+(3/2)−(5/2)=−1$
	$s o l u t i o n 1 {(1)}^{α} = 1$ $2 x^{2} + 5 x + 1 = 1 x (2 x + 5) = 0$ $x = 0 x = - \frac{5}{2}$ $You can't use 'macro parameter character #' in math mode$ $2 x - 3 = 0 x = \frac{3}{2}$ $s o l u t i o n 3 {(- 1)}^{2 n} = 1$ $2 x^{2} + 5 x + 1 = - 1$ $2 x^{2} + 5 x + 2 = 0$ $2 [{(x + \frac{5}{2})}^{2} - \frac{25}{4} + \frac{8}{4}] = 0$ $2 [{(x + \frac{5}{2})}^{2} - \frac{17}{4}] = 0$ $2 [(x + \frac{5}{2} - \frac{\sqrt{17}}{2}) (x + \frac{5}{2} + \frac{\sqrt{17}}{2})] = 0$ $x_{1} = - \frac{5}{2} + \frac{\sqrt{17}}{2} a n d x_{2} = - \frac{5}{2} - \frac{\sqrt{17}}{2}$ $t h e n x_{1} a n d x_{2} a r e n o t p e e r$ $s o i n c a s e 3 t h e r e i s n o t s o l u t i o n$ $S = {0 . \frac{3}{2} . - \frac{5}{2}}$ $w e h a v e 0 + \frac{3}{2} - \frac{5}{2} = - 1$
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