Question

Solve the following equations for x. Identify any extraneous solutions. Show all work.

-4√x+9=20

Answer 1

It isn't clear if the 9 is also under the radical..

Answer 2

I disagree.

Answer 3

\[-4\sqrt{x+9} = 20\]\[\implies \sqrt{x+9} = -5\]

And the square root cannot be negative, so there are no solutions.
And if the problem was:
\[-4\sqrt{x} + 9 = 20\]\[\implies -4\sqrt{x} = 11\]\[\implies \sqrt{x} =-\frac{11}{4}\]

Which also has no solutions.

Answer 4

zero displacement

Answer 5

x=16 is a valid solution. \[(-4)(-5)=20\]Or,\[-4\sqrt{25}=-4\sqrt{16+9}\]

Answer 6

So lets see:

\[-4\sqrt{x+9} = 20\]\[x = 16 \implies -4\sqrt{16 + 9} = 20\]\[\implies \sqrt{25} = -5\]\[\implies 5 = -5\]

Yeah, nope.

Answer 7

Even your own version isn't right because you are still equating -5 with sqrt(25). They aren't equal.

Answer 8

\[\sqrt{25}=\pm5\] as in:\[\sqrt{(-5)^2}=\sqrt{5^2}=\sqrt{25}\]

Answer 9

There is so much wrong with that...

Answer 10

sqrt25 does indeed equal -5

Answer 11

\[\sqrt{25} = 5\] Always.

Answer 12

I will grant you that \[\sqrt{25} = \sqrt{(-5)^2}\]But that's because \[ \sqrt{(-5)^2} = 5\]

Answer 13

You cannot take the square root of a number and get a negative.

Answer 14

Yes, you can.

Answer 15

Show me one number 'q' such that sqrt(q) < 0

Answer 16

http://en.wikipedia.org/wiki/Square_root

Answer 17

I know the definition of the square root. Did you come up with a q?

Answer 18

Every positive real number has TWO square root; one positive and one negative. So yeah..q is true for all positive numbers on the real number line. Pick one.

Answer 19

From the wikipedia article you quoted..:

Every non-negative real number x has a unique non-negative square root, called the principal square root, denoted by a radical sign as \( \sqrt{x}\).

So yes, there are postive and negative square roots, the \(\sqrt{x}\) is defined to be the positive one.

Answer 20

So you can certainly say that the equation:
\[x^2 = 3\]Has solutions \(x = \pm \sqrt{3}\)

But there are no solutions for which \(\sqrt{x} = q\) such that \(q < 0\)

Answer 21

Sorry had to clarify that the last equation wasn't related to the first two.

Answer 22

The square root function is the inverse function of the exponential function. Starting with the exponential function, we have: \[f(x)=x^2\]This function has domain and range:\[x \in \mathbb{R} ; y \ge0\]Now we find its inverse in the usual way:\[x=y^2\]and,\[x=\sqrt{y} \]and,\[y=\sqrt{x}\]The domain and range of an inverse function are swapped with the original function. i.e., the domain the square root function is the range of the exponential function:\[x \ge0\]No biggie, we still agree on this. The range of the square root function is the domain of the exponential function:\[y \in \mathbb{R}\]This obviously includes negative solutions. The domain and range are swapped for inverse functions because an inverse functions "undo" one another.

Answer 23

Good debate. I doubt my last one convinced you though. Off to bed..to be continued...

Answer 24

Only one to one functions have inverse functions.

Answer 25

This is why \(\sqrt{x} = q\) is defined to be the positive number q such that \(q^2 = x\).

Answer 26

Otherwise one input (25 for example) would have many ouputs (5 and -5). Functions don't behave this way. For one input there is one and only one output. So the function \(\sqrt{x}\) is defined to yield the positive value only. If you want a negative, you put a negative out in front of it (\(-\sqrt{x}\)) Or if you want both you can have both. \(\pm\sqrt{x}\)

Answer 27

It looks like I'm wrong on this one.
http://www.wolframalpha.com/input/?i=-4 √%28x%2B9%29%3D20+