Question

Please help, due today.

Which equation could be solved using this application of the quadratic formula?

Answer 1

Quadratic formula:
`x=(-b+/-√(b^2-4ac) )/(2a)` [ +/- is ± ]

You can find the value of a, b, c --> ax² + bx + c = 0

`x=(-2+/-√(2^2-4(1)(-4)) )/(2(1))` Since this is not simplified, you can find a, b, c:

a = 1

b = 2

c = -4

A.) x² + 1 = 2x − 3 Make the equation into ax² + bx + c = 0. Subtract 2x on both sides, and add 3 on both sides to set the equation equal to 0

x² + 1 - 2x + 3 = 2x - 2x - 3 + 3

x² - 2x + 4 = 0

a = 1

b = -2

c = 4 This is not the answer because b = 2 not -2, and c = -4 not 4

B.) x² - 2x − 1 = 3 Subtract 3 on both sides to set the equation = 0

x² - 2x - 4 = 0

a = 1

b = -2

c = -4 This is not the answer because b = 2 not -2

C. x² + 2x - 1 = 3 Subtract 3 on both sides to set the equation = 0

x² + 2x - 4 = 0

a = 1

b = 2

c = -4 This is your answer

D. x² + 2x - 1 = -3 Add 3 on both sides to set the equation = 0

x² + 2x + 2 = 0

a = 1

b = 2

c = 2 This is not the answer because c = -4 not 2