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

Answer 1

Kennedy Adams

#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#

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Answer 2

Hazel Anglin

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#

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Answer 3

Hazel Anglin

To solve the equation ( x^2 + 3 - 4x = 0 ):

Rearrange the equation to the standard quadratic form: ( x^2 - 4x + 3 = 0 )
Factor the quadratic expression: ( (x - 3)(x - 1) = 0 )
Set each factor equal to zero and solve for ( x ): ( x - 3 = 0 ) or ( x - 1 = 0 )
Solve for ( x ) in each equation: ( x = 3 ) or ( x = 1 )

So, the solutions are: ( x = 3 ) or ( x = 1 )

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.