2xsquared +7x+3=0

Question

2x²+7x+3=0

How do you figure this out

Answer 1

2*3 = 1*6, and 1+6 = 7, the coefficient of 7x.

So, you can split 7x into x+6x,

2x2+7x+3 = 0

=> 2x2+6x + x+3 = 0

=> 2x(x+3) + (x+3) = 0

=> (2x+1)(x+3) = 0

Answer: x = -1/2, -3

Answer 2

Hi Brandon;

2x²+7x+3=0

Let's factor...

For the FOIL...

FIRST must be (2x)(x)

LAST must be (3)(1)

OUTER and INNER must be 2(either 3 or 1) + (either 3 or 1)

So...

(2x+1)(x+3)=0

FOILed, this is...

2x²+6x+x+3=2x²+7x+3

 

(2x+1)(x+3)=0

Either parenthetical equation can =0

(2x+1)=0 and/or (x+3)=0

2x=-1

x=-1/2     and x=-3

 

 

 

 

Answer 3

2x^2+7x+3=0

 

2=1*2 and 3=1*3 (these are the only factorizations)

 you have 1 and 2

 you have 1 and 3

 

one number from the 1 and 2 multiplies one number from the 1 and 3

the remaining numbers also multiply

we need to get 7

1*1=1 and 2*3=6, 1+6=7

 

(2x+?)(x+?) is what we have so far

remember the 2 must multiply the 3 and the 1 must multiply the 1

so we have (2x+1)(x+3)=0

                  2x+1=0 and x+3=0

                  2x=-1, x=-1/2  and x=-3

 

Answer 4

There are two ways to solve this. By factoring and by using the quadratic formula.  

 

In both observe the form of the quadratic, ax²+bx+c

 

To factor:

Multiply a×c, in this case 2×3, which is 6

Break up the bx term to 6x + x giving you

2x²+6x+x+3

 

Group the terms as [2x²+6x]+[x+3] and factor each group (if there are no factors, factor out a 1)

 

2x(x+3) + 1(x+3)

 

Now factor out the (x+3) from both giving

 

(x+3)(2x+1)=0

Setting both factors to 0 and solve gives the solution

x=-3 and x=-½

 

Using quadratic formula:

a=2 b=7 c=3

 

x= (-b±√(b²-4ac))/2a

x= (-7+√(49-24))/4 and x= (-7-√(49-24))/4

x= (-7+√25)/4 and x=(-7-√25)/4

x= (-7+5)/4 and x=(-7-5)/4

x= -2/4 and x= -12/4

x= -1/2 and x=-3