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

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Answer 1 · 2016-07-28T23:02:57

#x = 3 or x=1#

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#

Answer 2 · 2016-07-29T04:53:46

1 and 3

#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#