How do you solve #x^2+3-4x=0#?

2 Answers
Jul 28, 2016

Answer:

#x = 3 or x=1#

Explanation:

The first approach with quadratic equations is to see if we can find factors. Use #ax^2 + bx + c = 0#

#x^2 -4x +3 = 0#

Find factors of 3 which add up to 4.
The signs are the same, both minus.

#(x-3)(x-1) = 0#

Either of the factors can be 0.so:
#x = 3 or x=1#

Jul 29, 2016

Answer:

1 and 3

Explanation:

#y = x^2 - 4x + 3 = 0#
Since a + b + c = 0, use shortcut. One real root is 1 and the other is #c/a = 3#