Solve the quadratic equation 3x2 + 2x – 16 = 0 using the quadratic formula.

Question

 Solve the quadratic equation 3x2 + 2x – 16 = 0 using the quadratic formula. 

Answer 1

3X^2 + 2X - 16 = 0             3( -16) = -48    8 - 6 = 2   -48 = 8 * ( -6)  This quadratic has linear factors of

                                           rational numbers .

 

   3 X^2 - 6X + 8X - 16 =

 

     3X( X -2 ) + 8 ( X -2 ) = 0

 

      ( X -2 ) ( 3X + 8 ) =0

 

      X - 2 = 0   X =2

 

       3X + 8 = 0   X = -8/3

 

 

 

 

 

 

Answer 2

Hi Pamela;

3x² + 2x – 16 = 0

quadratic formula is...

x=[-b+/-√(b²-4ac)]/(2a)

4ac=(4)(3)(-16)=-192

x=[-2+/-√(2²--192)]/[(2)(3)]

x=[-2+/-√(4+192)]/6

x=(-2+/-√196)/6

x=(-2+/-14)/6

x=(-2+14)/6......x=(-2-14)/6

x=12/6..............x=-16/6

x=2...................x=-2 4/6

.........................x=-2 2/3

 

Let's check our results and factor the equation...

For the FOIL...

FIRST must be (3x)(x)=3x²

OUTER and INNER must add-up to 2x.

LAST must be (16)(1) or (1)(16) or (8)(2) or (2)(8) or (4)(4) and only one number must be negative to render the sum of +2 and product of -16.

(3x+8)(x-2)=0

Let's FOIL...

FIRST...(3x)(x)=3x²

OUTER...(3x)(-2)=-6x

INNER..(8)(x)=8x

LAST...(8)(-2)=-16

3x²-6x+8x-16=0

3x²+2x-16=0

(3x+8)(x-2)=0

Either or both parenthetical equation(s) must equal zero...

(3x+8)=0.........(x-2)=0

3x=-8..............x=2

x=-8/3

x=-2 2/3

Answers are identical.

Answer 3

If ax^2 + bx + c = 0, then

 

x = (-b ± √(b^2-4ac))/(2a).

 

3x^2 + 2x – 16 = 0

 

a = 3, b = 2, c = -16

 

x = (-(2) ± √((2)^2-4(3)(-16)))/(2(3))

 

x = (-2 ± √(4+4(3)(16)))/(2(3))

 

x = (-2 ± √(4(1+(3)(16))))/(2(3))

 

x = (-2 ± √(4(49))/(2(3))

 

x = (-2 ± 2(7))/(2(3))

 

x = (-1 ± 7)/3 = -8/3, 2

check:

3(-8/3)^2 + 2(-8/3) – 16 =? 0
3(64/9) - 16/3 – 16 =? 0
64/3 - 16/3 – 3*16/3 =? 0
64/3 - (1+3)16/3 =? 0
64/3 - 4*16/3 = 0 √

3(2)^2 + 2(2) – 16 =? 0
3*4 + 4 – 16 =? 0
12 + 4 – 16 = 0 √