SOLUTION: Please help me to solve this problem {{{y^2 =x^2 - 4x + 3}}}

Algebra.Com
Question 421872: Please help me to solve this problem
Found 2 solutions by Gogonati, Theo:
Answer by Gogonati(855)   (Show Source): You can put this solution on YOUR website!

We can make some transformations at initial equation:
Add on both sides of equation.
The final equation is the equation of hyperbola with center
the point (2,0).

Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
y^2 = x^2 - 4x + 3

set y^2 = 0 and you get:

x^2 - 4x + 3 = 0 which leads to (x-1)*(x-3) = 0 which leads to x = 1 and x = 3.

these are the roots of the quadratic equation.

a graph of that equation is shown below:



this, however, is the graph of y = ....., not y^2 = .....

the graph of y^2 = x^2 - 4x + 3 is shown below:



I'm just guessing because I've never seen a problem such as this before, but it appears that the roots are the same, except the graph of the equation is different.

this stands to reason because the equations are different after you solve for y.

y = x^2 - 4x + 3

versus

y = +/- sqrt(x^2 - 4x + 3)

in both cases, x is the independent variable and y is the dependent.

you select values of x for plotting your graph and then you calculate what the value of y would be for those values of x based on the equation.

in the equation of y^2 = x^2 - 4x + 3, the zone between x = 1 and x = 3 does not show on the graph most likely because the value of y is equal to the square root of a negative number which is not real, therefore can't be graphed on a graph that is dealing with real numbers only.

for example, when x = 2, y^2 = x^2 - 4x + 3 becomes 4 - 8 + 3 = -4 + 3 = -1

y = +/- sqrt(-1) is not a real number, therefore can't be graphed.

does this answer your question or were you looking for something else?



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