How do you solve $- x^{2} + 2 x - 1 = 0$ $-x^2+2x-1=0$ ?

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How do you solve $- x^{2} + 2 x - 1 = 0$ $-x^2+2x-1=0$ ?

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Answer 1 · 2018-05-29T06:41:52

x=1

Explanation:

IT IS EASY!!
$- x^{2} + 2 x - 1 = 0$ $-x^2+2x-1=0$ { * COMPARING WITH *
$a x^{2} + b x + c$ $ax^2+bx+c$
** we see **
$a = - 1, b = 2, c = - 1$ $a=-1,b=2,c=-1$
by splitting the middle term....
* WE GET,
PRODUCT= $c a = - 1 \cdot - 1 = 1$ $ca = -1*-1=1$
SUM = $2$ $2$ *
therefore
$- x^{2} + x + x - 1 = 0$ $-x^2+x+x-1=0$
$- x (x - 1) + (x - 1) = 0$ $-x(x-1)+(x-1)=0$
$(- x + 1) (x - 1) = 0$ $(-x+1)(x-1)=0$
hence
x=1................................. hope u get it
** * CHEERS !!:D * **

Answer 2 · 2018-05-29T06:52:04

$x = 1$ $x=1$ .

Explanation:

This equation can be solved by factorization.

$- x^{2} + 2 x - 1 = 0$ $-x^2 + 2x - 1 = 0$

$- 1 (x^{2} - 2 x + 1) = 0$ $-1(x^2 -2x + 1) = 0$

$x^{2} - 2 x + 1 = 0$ $x^2 -2x + 1 = 0$

$x^{2} - x - x + 1 = 0$ $x^2 - x - x + 1 =0$

$x (x - 1) - 1 (x - 1) = 0$ $x( x-1 ) -1( x-1 ) =0$

$(x - 1) (x - 1) = 0$ $( x-1 )( x-1 ) = 0$

Now,
$x - 1 = 0$ $x-1 = 0$

Therefore, $x = 1$ $x=1$ .

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How do you solve $- x^{2} + 2 x - 1 = 0$ $-x^2+2x-1=0$ ?

2 Answers

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