How do you find the roots, real and imaginary, of y= x^2 - 3x + 44-(3x-1)^2 y=x2−3x+44−(3x−1)2 using the quadratic formula?

1 Answer
Gagan Batra
Aug 9, 2017

You should know the 2 formulas:
1.) (a - b)^2 = a^2+b^2-2ab(a−b)2=a2+b2−2ab
2.) Quadratic Formula: (-b+-sqrt(b^2-4ac))/(2a)−b±√b2−4ac2a

Explanation:

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

y=x^2-3x+44-(3x-1)^2y=x2−3x+44−(3x−1)2
y=x^2-3x+44-(9x^2+1-6x)y=x2−3x+44−(9x2+1−6x)

Now solvng it further:
y=x^2-3x+44-9x^2-1+6xy=x2−3x+44−9x2−1+6x
y=-8x^2+3x+43y=−8x2+3x+43

Step 2.) Now applying Quadratic formula:

(-b+-sqrt(b^2-4ac))/(2a)−b±√b2−4ac2a
x= (-3+-sqrt(3^2-4(43)(-8)))/(2(-8)x=−3±√32−4(43)(−8)2(−8)
x= (-3+-sqrt(9+1376))/(-16)x=−3±√9+1376−16
x= (-3+-sqrt(1385))/(-16)x=−3±√1385−16
x= (-3+-37.21)/(-16)x=−3±37.21−16

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

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

That's all, Thank You.

Related questions
See all questions in Quadratic Formula
Impact of this question
1838 views around the world
You can reuse this answer
Creative Commons License