How do you find the roots, real and imaginary, of #y=x^2 - 3x + 2 # using the quadratic formula?

1 Answer
Dec 9, 2015

Answer:

You can find all the roots of #y = x^2 - 3x + 2# by "plugging in" values from the equation into the quadratic formula to find that the roots are #(2, 0)# and #(1, 0)#.

Explanation:

The quadratic formula is:

#x = (-b +- sqrt(b^2 - 4ac))/(2a)#

The equation representing any given quadratic, or parabola, can be:

#y = ax^2 + bx + c#

So all we need to do is assign values of our equation to their respective variables, and then substitute them into the quadratic formula to solve. In the equation #y = x^2 - 3x + 2#:

#a = 1#, #b = (-3)#, and #c = 2#

#x = (-b +- sqrt(b^2 - 4ac))/(2a)# Given Quadratic Formula

#x = (-(-3) +- sqrt((-3)^2 - 4(1)(2)))/(2(1))# Substitute

#x = (3 +- sqrt(9 - 4(2)))/(2)# Simplify

#x = (3 +- sqrt(9 - 8))/(2)# Simplify Some More

#x = (3 +- sqrt(1))/(2)# Simplify A Little More

#x = (3 +- 1)/(2)# Simplify Even More

#x = (3 + 1)/2#, OR #x = (3 - 1)/2# Isolate Both Possible Roots

#x = 2# OR #x = 1# Simplify (For the Last Time..)

This means that the roots, or x-intercepts, of our quadratic equation are at the points: #(2, 0)# and #(1, 0)#.