(2x+1)(x+2)=5x+2?

Question

I have to solve (2x+1)(x+2)=5x+2 for x. Somebody please help with work?

Also need help with:

(x-2)/(2x+1) > 0 solve for x

Answer 1

sq=squared
(2x+1)(x+2)=5x+2

2xsq+4x+x+2=5x+2 timesd left side
2xsq+5x+2=5x+2
2xsq+5x+2-5x=2 put 5x to other side
2xsq+2=2 divided by2
xsq+1=1
ya xsq = 0?

Answer 2

I highly believe you should use FOIL, which stands for First, Outside, Inside, Last, on the left hand side of this equation before you solve for x.
First Multiply 2x by x: (2x+1)(x+2)=5x+2
2x times x=2x squared. This is the first step which is multiplying the first two terms.
Second Multiply 2x by the 2 in the right parentheses. (2x+1)(x+2)=5x+2

2x times 2= 4x This is the second step which is multiplying the outer terms.
Third multiply the 1 in the left parentheses by the x in the right parentheses.
(2x+1)(x+2)=5x+2

1 times x=x This is the third steps which is multiplying the two inner terms.
Fourth multiply 2 by 1. (2x+1)(x+2)=5x+2 2 times 1=2

This is the fourth step which is multiplying the two last terms.
After completing these steps you are left with 2x squared + 4x+x+2=5x+2
Next combine like terms: 2x squared + 5x+2=5x+2
Now isolate zero on one side and solve for x
subtract 5x+2 from both sides
You are left with 2x squared = zero
Divide 2 by both sides and you are left with x^2=0
Take the square root of both sides and you are left with x=0

CHECK
Check by substituting in zero for x into the orginal equation (2(0)+1)(0+2)=5(0)+2
(0+1)(2)=0+2
(1)2=2
2=2
Eureka! Your equation is solved.

Answer 3

ok let's start off with (2x+1)(x+2)
You can use FOIL to do this because it's a binomial (2 sets of polynomials.)
FOIL
F - Multiply the 2 fronts
O - Multiply the 2 on the outside
I - Multiply the ones inside
L - Multiply the ones that are last.

You should end up with 2x^2 + 4x + x + 2
Simplify that and you get
2x^2 + 5x + 2
Now let's bring out the full equation
2x^2 + 5x + 2 = 5x + 2
Subtract 2 from both sides, so they cancel out.
2x^2 + 5x = 5x
Subtract 5x from both sides, and they also cancel out.
You're left with
2x^2.
so
x = 0

Answer 4

(2x+1)(x+2)=5x+2
=>x*(2x+1)+2*(2x+1)=5x+2
=>2x^2 + x +4x+2=5x+2
=>2x^2 +5x+2=5x+2
=>2x^2 =0
=>x^2 =0
=>x =0



Additonal question
(x-2)/(2x+1) > 0
=>(x-2) > 0 [multiplying both sides by (2x+1)]
=>x > 2 [adding 2 to both sides]

Answer 5

After foil on the left side you get 2x^2+5x+2=5x=2, then move 5x on right to left side and 2 on left side to right. Then you have 2x^2=0 therefore x must equal 0.

Answer 6

you need to combine (2x+1)and(x+2). that is 2x^2+5x+2=5x+2.
Then you cancel out the 5x on both sides ==> 2x^2+2=2.
Subtract 2 from both sides. ===> 2x^2=0, therefore x=0

Answer 7

simplify
2x2 + 5x + 2 = 5x +2
subtract 5x +2 from both sides3
2x2 = 0
x = 0

Answer 8

1.
Expand LHS & RHS to give
2x^2+5x+2 = 5x+2
Therefore 2x^2=0
x=0;

2.
(x-2)/(2x+1) > 0 solve for x
If 2x+1>0, then multiply both sides by 2x+1 to give
(x-2)>0
or x>2

if 2x+1<0
then multiply both sides by 2x+1 to give
(x-2)<0 (change inequality when multiplied by negative number)
x<2, but since it was presumed that x<0, therefore x=0 to x=2 will be excluded to give x<0.
Furthermore, to avoid singularity, x<>-1/2
So the solution set is
x<-1/2; -1/22

Answer 9

(2x + 1)(x + 2) = 5x + 2
2x^2 + 4x + x + 2 = 5x + 2
2x^2 + 5x + 2 = 5x + 2
2x^2 = 0
x^2 = 0
x = 0

ANS : x = 0

------------------------

(x-2)/(2x+1) > 0 solve for x

x - 2 > 0
x > 2

since (-1/2) would make this problem undefined

x < (-1/2)
or
x > 2

Answer 10

(2x+1) (x+2) = 5x+2
2x*x + 5x + 2 = 5x + 2
2x*x = 0
x =0.

In general, to solve a quadratic equation of the form:
ax*x + bx +c = 0

x = (-b +/- sqrt(b*b - 4ac) )/2a