This is a radical equation and a linear equation
When we get to the part where we solve x intercept in the quadratic equation, why does it work that we can plug the x values back into our original equations that we can find their intercepts?
I want to know the how, I don’t want to memorize
I’m not sure if I understand you right, when you say “plugging in the x values back into our original equations”, what are the x values you are taking about?
Please elaborate on this topic. Your question is unclear.
I don’t believe that it actually works like that. If you plug in x = -4 for the left side you get positive 1. And if you plug in -1 to the right you get positive 4.
First I think you need to understahd how a function works. Here is a simple linear function
f(x) = 6x + 2 or y = 6x + 2
In a function you feed it an x value and it spits out a y value. If we have a y value of 0 then that means we solved for an x intercept. Using this knowledge we can solve for a functions x intercepts.
0 = 6x +2
-2 = 6x
× = -1/3
Thats the extraneous value, the answer is -1
I understand how functions work, I just want the WHY the x intercept of the quadratic function is the same as the intercept of the radical and linear function
What, all x intercepts are the same? Its just when the y value is 0. There is no difference in them except the way you do the algebra to solve them
I believe that is is just a coincidence, also how is that an extraneous value. Plug in -4 and you get
(sqrt(-4+5))^2. Then add (sqrt(1))^2 sqrt and squaring it cancel out and have no effect, so you get 1.
Extraneous means it is a solution, but is invalid for one reason or another. In this case, -4 doesn’t work because x+3 does not intercept sqrt(x+5) at x = -4. I hope that makes sense.
I believe that it does. Pluging it in to the left side you get positive 1. On the right side you also get positive 1. Right Side: (-4+3)^2 (-1)^2 1.
Expand, FOIL, Simplify.
(x+3)(x+3)
x^2+3x+3x+9
x^2+6x+9
(x+3)^2 = x^2 + 6x + 9
30lettersrequired
When you square a binomial you have to multiply all terms by eachother. xx = x^2, x3 = 3x, x^3 = 3x, 3*3 = 9. Take the sum and you get x^2 + 6x + 9.
But it is invalid because 1 does not equal -1. sqrt(-4 + 5) = -4 + 3. Simplifies to 1 = -1. Inequal in that case.
But the right side is not -4 + 3, it is (-4 + 3)^2,
(-1)^ = 1
You are adding in an unnecessary squaring of the left side. Square root of 1 is just 1, there is no need to square both sides. This is why it is a solution, due to square roots being ±, however, it is not a solution for the primary equation. You can see it graphically:
There is no intersection at x = -4. Graphing an equation can tell you pretty much anything you need to know about it.