Question

70 points! someone please help 3

Answer 1

Part A

a = 1

b = 3

c = -4

Part B

`x = ( -3 \pm √(3^2 - 4(1)(-4)))/( 2(1) )`

Part C

`\mathrm{x = ( -3 + √(25))/( 2 ) :and\: x = ( -3- √(25))/( 2 )}`

Part D

Option B: x = 1 and x = -4

Explanation:

The standard form of the quadratic equation is
`ax^2 + bx + c=0`

1.The given equation is
`x^2 + 3x - 4 = 0`

Part A

Comparing the given equation with the generalized standard equation we see that

`a` = coefficient of
`x^2` = 1

`b`= coefficient of
`x = 3`

`c`= constant term =
`-4`

Part B.

The quadratic formula for the roots of the equation is:

`x = ( -b \pm √(b^2 - 4ac))/( 2a )`

Plugging in values for a, b and c from Part A:

`x = ( -3 \pm √(3^2 - 4(1)(-4)))/( 2(1) )`

Part C

`\mathrm{x = ( -3 + √(25))/( 2 ) :and\: x = ( -3- √(25))/( 2 )}`

part D
Simplifying we get

`x = ( -3 + 5\, )/( 2 ) = (2)/(2) = 1\\\\x = ( -3 - 5\, )/( 2 ) = (-8)/(2) = -4\\\\`

Correct answer for Part C:
B. x = 1 and x = -4