I am having a problem finding the x intercepts.
I have found the Standard form of the Quadratic (Vertex form) -(x+2)^2 + 8 [by the way, does is matter if the 8 is in front of the minus sign or in the back?]
y intercept = (0,4)
To find the x intercept I did the following:
-(x+2)^2+8
0 = -(x+2)^2+8
-8 = -(x+2)^2
(-8)/(-1) = (-(x+2)^2) / -1
8 = (x+2)^2
Square root both sides:
square root 2, square root 4 = x+2
Now this is where I am stuck
The answer key in the book states that the x intercept is +2, square root of 2 (minus) 2
and -2,-2 the square root of 2
Yet, after you simplify the radicals doesn't the 2 come directly out in front?
Other questions to clarify my understanding:
Do we not place the +/- in front of the simplified radical?
Does we not place the -2 in front of the + and - sign?
This is Pre-Calculus class, thus my algebra may be flawed. Please indicate what I am failing to understand. Please explain as if I am a 5 year old.
Thanks in advance.
$\endgroup$ 33 Answers
$\begingroup$I will just use the quadratic formula for this. $$f(x)=-x^2-4x+4$$ We want $f(x)$ to be equal to $0$ (because that is where the x-intercepts are). So now our equation becomes $$0=-x^2-4x+4$$ Remember, the quadratic formula states that for any equation $ax^2+bx+c=0$ (where $a\neq 0$), the roots are $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ We can apply this formula to our equation. In our equation, $a=-1$, $b=-4$, and $c=4$. $$x=\frac{-(-4)\pm\sqrt{(-4)^2-4(-1)(4)}}{2(-1)}$$ $$x=\frac{4\pm\sqrt{16+16}}{-2}$$ $$x=\frac{4\pm\sqrt{32}}{-2}$$ $$x=-\frac{4\pm4\sqrt{2}}{2}$$ $$x=-2\pm 2\sqrt{2}$$ Therefore the x-intercepts are $-2+2\sqrt{2}$ and $-2-\sqrt{2}$. Hope I helped!
$\endgroup$ 1 $\begingroup$The $x$ intercept is a function defined as the points where it cuts the $x$ axis, that is , all point of the form $(\alpha,0)$ where $\alpha$ is a root of the function.
Solving for the root,
$$-x^2-4x+4=0$$
$$x^2+4x-4=0$$
Using the quadratic formula,
$$x=\frac{-4\pm \sqrt{16+16}}{2}$$
$$x=-2+2\sqrt2, -2-2\sqrt{2}$$
Thus the $x$ intercepts are,
$$(2\sqrt2 -2,0)\text{ and } (-2\sqrt{2}-2,0)$$
Note that, you have written,
$$(x+2)^2=8\implies x+2=\sqrt{4}\sqrt{2}\tag{*}$$
(*) I assume that is what you mean by square root four square root 2
Mistake is, $x^2=y^2 \not \rightarrow x=y$, since $x=-y$ is valid as well.
Also, there is a way to continue, your work.
You write, $$(x+2)^2=8$$
which is right.
$$x+2=\color{red}{\pm}2\sqrt{2}$$
$$x=-2\pm2\sqrt2$$
which is the same answer.
$\endgroup$ 6 $\begingroup$Let's start from the point where you got stuck:-
$$8=(x+2)^2$$
If you take the square root of both sides you obtain
$$\pm\sqrt{8}=(x+2)$$
The $\pm$ sign means that the square root can be either positive or negative - there are two possible answers.
Now we have the following:
$$\pm\sqrt{8}=\pm\sqrt{2\times4}=\pm\sqrt{2}\times\sqrt{4}=\pm\sqrt{2}\times2=\pm2\sqrt{2}$$
So you have $$\pm2\sqrt{2}=(x+2)$$
Putting all the knowns on the left hand side, we have
$$\pm2\sqrt{2}-2=x$$
So the answer is
$$x=\pm2\sqrt{2}-2$$
which is the same as saying that there are two answers:-
$$x=2\sqrt{2}-2$$ $$x=-2\sqrt{2}-2$$
Thus the two intercepts are at $(2\sqrt{2}-2,0)$ and $(-2\sqrt{2}-2,0)$.
These match the answer key in your book, because $-2-2\sqrt{2}$ is the same as $-2\sqrt{2}-2$.
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