How do you solve $x^2-2x-1=x+3$ ?

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Answer 1 · 2018-04-13T05:15:13

$x=4$
$x=-1$

First, get $0$ on one side.
$(x^2-2x-1)$ $-(x+3)$ = $x+3$ $-(x+3$ )

$x^2-3x-4=0$

Next, factor this equation.
$(x-4)(x+1)=0$
Therefore,
$(x-4)=0$
$x=4$
or
$(x+1)=0$ ,
$x=-1$

You can verify this using FOIL:
= $x^2+x-4x-4$
= $(x^2-3x-4)$

Answer 2 · 2018-04-13T05:22:02

$x=-1$
$x=4$

$x^2-2x-1=x+3$

$x^2-2x-1-x-3=0$

$x^2-3x-4=0$

$x^2-4x+x-4=0$

$x(x-4)+1(x-4)=0$

$(x+1)(x-4)=0$

$x+1=0$
$x=-1$

$x-4=0$
$x=4$

Answer 3 · 2018-04-13T05:22:45

$x=-1,x=4$

Consider the given equation
$x^2-2x-1=x+3$

Subtract $x+3$ from both sides (Or bring all terms to the left-hand side)

$x^2-2x-1-x-3=x+3-x-3$

Thus,
$x^2-2x-1-x-3=0$

Add like terms
$x^2-3x-4=0$

Split middle term so as to obtain $(-4)$ as product and $(-3)$ as sum
$x^2+x-4x-4=0$

Take common factors out from paired terms
$x(x+1)-4(x+1)=0$

Then,
$(x+1)(x-4)=0$
$x+1=0$ or $x-4=0$
$x=-1$ or $x=4$

How do you solve x^2-2x-1=x+3x^2-2x-1=x+3?