Question

How do you find the roots, real and imaginary, of `y= x^2 - 3x + 44-(3x-1)^2` using the quadratic formula?

Answer 1

You should know the 2 formulas:
1.) `(a - b)^2 = a^2+b^2-2ab`
2.) Quadratic Formula: `(-b+-sqrt(b^2-4ac))/(2a)`

Explanation:

Step 1.) Open the bracket term in question by formula no.1 and solve it further:

`y=x^2-3x+44-(3x-1)^2`
`y=x^2-3x+44-(9x^2+1-6x)`

Now solvng it further:
`y=x^2-3x+44-9x^2-1+6x`
`y=-8x^2+3x+43`

Step 2.) Now applying Quadratic formula:

`(-b+-sqrt(b^2-4ac))/(2a)`
`x= (-3+-sqrt(3^2-4(43)(-8)))/(2(-8)`
`x= (-3+-sqrt(9+1376))/(-16)`
`x= (-3+-sqrt(1385))/(-16)`
`x= (-3+-37.21)/(-16)`

Case 1:
when the sign is positive in between:
`x= (-3+37.21)/(-16)`
`x= (34.21)/(-16)`
`x=- 2.13`

Case 2:
when the negative is positive in between:
`x= (-3-37.21)/(-16)`
`x= (-40.21)/(-16)`
`x= 2.51`

That's all, Thank You.