Find x: (3x + 2)(2x + 1) = 1?

6x^2+7x+2=1
6x^2+7x+1=0
(6x+1)(x+1)=0
x=-1/6 or -1

3x(2x+1)+2(2x+1)=1
6x^2+3x+4x+2=1
6x^2+7x+1=0
6x^2+6x+x+1=0
6x(x+1)+1(x+1)=0
(6x+1)(x+1)=0
(6x+1)=0 (x+1)=0
x=-1/6 x=-1

(3x + 2)(2x + 1) = 1
3x*2x + 2*2x + 3x*1 + 2*1 = 1
6x^2 + 4x + 3x + 2 = 1
6x^2 + 7x + 2 - 1 = 0
6x^2 + 7x + 1 = 0
(6x + 1)(x + 1) = 0

6x + 1 = 0
6x = -1
x = -1/6

x + 1 = 0
x = -1

∴ x = -1/6 , -1

6x² + 7x + 2 = 1
6x² + 7x + 1 = 0
(6x + 1)(x + 1) = 0
x = - 1/6 , x = - 1

hi helen !

(3x + 2) (2x + 1) = 1
6x ^ 2 + 3x + 4x +2 = 1
6x ^ 2 + 7x +2 -1 = 0
6x ^ 2 + 7x + 1 = 0

now we have to find 2 no's whose product equals to 6 and addition of those 2 no's equals 7

so the 2 no's are 6 & 1

6x ^2 + 6x + 1x + 1 = 0
6x ( x + 1) + 1 ( x + 1 ) = 0
( 6x + 1 ) ( x + 1 ) = 0
(6x +1) = 0 OR (x+1)=0
x = [-1/6] or x = [ -1 ]

First you solve the equation in a normal way:

(3x+2)(2x+1)=1
6x^2+3x+4x+2=1
6x^2+7x+2-1=0
6x^2+7x+1=0

In this case you cannot factorize and for this reason you use the next formula:

x=-b PLUS or MINUS sqrt b^2-4ac/2a

So:

-7 PLUS or MINUS sqrt 7^2-4*6*1/2*6
-7 PLUS or MINUS sqrt 49-24/12
-7 PLUS or MINUS sqrt 25/12
-7 PLUS or MINUS 5/12

From here we have the two solutions that guide us to know the terms of X:

First solution of X:

-7-5/12=-1

Second solution X:

-7+5/12=-1/6

That's it:

X is either -1 or -1/6

(3x+2)(2x+1)=1 <=> 3x+2=1 and 2x+1=1<=> x= -1/3 and x=0 IT'S WRONG.
(3x+2)(2x+1)=1<=>3x+2= -1 and 2x+1= -1<=> x= -1 and x= -1
IT'S RIGHT.
so (3x+2)(2x+1) <=> x= -1

firstly multiply out the brackets to get 6x2 + 3x + 4x + 2=1

this simplifies to 6x2 + 7x +1 = 0 (always set to zero to find the roots)

equation doesnt factorise simply so use (b+/- sqroot(b2-4ac))/2a

(ax2 +bx + c)

answer comes out as x=1/6, 1

Just expand it out

(3x + 2)(2x + 1) = 6x² + 4x + 3x + 2 = 6x² + 7x + 2
so
6x² + 7x + 2 = 1

6x² + 7x + 1 = 0

Solvew in the normal way