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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.

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  • $\begingroup$ The $x$ intercept is defined at the point, where the graph cuts the $x$ axis, i.e, the roots of the function. Do you know how to solve, $$-x^2-4x+4=0$$ $\endgroup$
    – Guy
    Apr 5, 2014 at 9:07
  • $\begingroup$ Square root of 8 is not 4 neither $\sqrt{4}$!!! $\endgroup$
    – barak manos
    Apr 5, 2014 at 9:10
  • $\begingroup$ Also, $x^2=y^2 \not \rightarrow x=y$ $\endgroup$
    – Guy
    Apr 5, 2014 at 9:12

3 Answers 3

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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!

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  • $\begingroup$ Thank you for your help, and for showing all the steps. $\endgroup$
    – Cetshwayo
    Apr 5, 2014 at 9:27
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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.

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  • $\begingroup$ I never stated that it equaled 8. Also, when you used the Quadratic formula, the 2 in the denominator is a -2, since a is -1. How did you get +2? $\endgroup$
    – Cetshwayo
    Apr 5, 2014 at 9:24
  • $\begingroup$ @Utvecklaochförenkla you did state it equals $8$, read again please(which is correct, you just cannot take the square root) Also, just multiply both sides by $-1$. Surely the roots of $f(x)$ and $-f(x)$ are the same. $\endgroup$
    – Guy
    Apr 5, 2014 at 9:27
  • $\begingroup$ Good enough. Miss-communication $\endgroup$
    – Cetshwayo
    Apr 5, 2014 at 9:29
  • $\begingroup$ @Utvecklaochförenkla I added a edit. You can continue from $(x+2)^2=8$. Your work up until that point is right. $\endgroup$
    – Guy
    Apr 5, 2014 at 9:30
  • $\begingroup$ Yes, I can see that. Thank You! $\endgroup$
    – Cetshwayo
    Apr 5, 2014 at 9:33
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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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  • $\begingroup$ As for the quadratic in Standard form. Why does it not matter if it is in front of the binomial or in the back, for example 8-(x+2)^2 = (x+2)^2 +8. Also, why mathematically does it not matter if the -2 is in the front or back of the the square root in the solution? Thanks in advance. $\endgroup$
    – Cetshwayo
    Apr 5, 2014 at 9:40
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    $\begingroup$ Looking at the quadratic in standard form $8-(x+2)^2=-(x+2)^2+8$ (take care with the signs). In terms of the solution, $-2\sqrt{2}-2$ is the same as $-2-2\sqrt{2}$ because $a+b$ is the same as $b+a$. An easier way to look at the solution is to express $-2-2\sqrt{2}$ as $(-2)+(-2\sqrt{2})$, which is the same as $(-2\sqrt{2})+(-2)$ - taking away the brackets, you end up with the answer key in you book. $\endgroup$
    – Alijah Ahmed
    Apr 5, 2014 at 9:47
  • $\begingroup$ Commutative property, Got it. Thanks for being such a great teacher. Really appreciate it.- $\endgroup$
    – Cetshwayo
    Apr 5, 2014 at 9:52
  • $\begingroup$ Your welcome, and well done for mentioning the commutative property. $\endgroup$
    – Alijah Ahmed
    Apr 5, 2014 at 10:02

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