The Purplemath Forums

by **santaclaus** on Mon Mar 09, 2009 7:18 pm

x² - square root of x = 4 - square root of x
x = 4
x= plus or minus 2

Actually, -2 does not satisfy the original equation. Explain the appearance of the extraneous solution. If both sides of the original equations are functions, what is happening here? -- don't understand what extraneous solution means.

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by **stapel_eliz** on Mon Mar 09, 2009 7:51 pm

santaclaus wrote:x² - square root of x = 4 - square root of x
x = 4
x= plus or minus 2

Actually, -2 does not satisfy the original equation. Explain the appearance of the extraneous solution. If both sides of the original equations are functions, what is happening here? -- don't understand what extraneous solution means.

To learn how to solve radical equations, what "extraneous" solutions are, and how to deal with them, try here. :wink:

by **santaclaus** on Mon Mar 09, 2009 8:21 pm

To learn how to solve radical equations, what "extraneous" solutions are, and how to deal with them, try here.
so, when I looked at the help area you told me to go to, I saw that you had to square both sides and then solve the equation: so, i tried: x⁴-2x+x=16-4x+x
then that came to x⁴ - 2x- 16. you can't reduce this anymore. so, what next? or did I make a mistake doing the math?

by **stapel_eliz** on Mon Mar 09, 2009 11:22 pm

santaclaus wrote:...when I looked at the help area you told me to go to, I saw that you had to square both sides and then solve the equation: so, i tried: x⁴-2x+x=16-4x+x

Your original equation was:

. . . . . $x^2\, -\, \sqrt{x}\, =\, 4\, -\, \sqrt{x}$

Squaring the left-hand side gives:

. . . . . $\left(x^2\, -\, \sqrt{x}\right)^2\, =\, \left(x^2\, -\, \sqrt{x}\right)\left(x^2\, -\, \sqrt{x}\right)$

. . . . . $=\, (x^2)^2\, -\, 2(x^2)(\sqrt{x})\, +\, (\sqrt{x})^2\, =\, x^4\, -\, 2x^2 \sqrt{x}\, +\, x$

Squaring the right-hand side gives:

. . . . . $\left(4\, -\, \sqrt{x}\right)^2\, =\, \left(4\, -\, \sqrt{x}\right)\left(4\, -\, \sqrt{x}\right)$

. . . . . $=\, (4)^2\, -\, 2(4)(\sqrt{x})\, +\, (\sqrt{x})^2\, =\, 16\, -\, 8\sqrt{x}\, +\, x$

(To learn how to multiply polyomials and polynomial-like expressions, try here. To learn how to work with radicals, try here.

To solve the resulting equation, you'll need to isolate the radicals on one side, and then square again.

. . . . . $x^4\, -\, 2x^2 \sqrt{x}\, +\, x\, =\, 16\, -\, 8\sqrt{x}\, +\, x$

. . . . . $x^4\, +\, x\, -\, 16\, -\, x\, =\, -8\sqrt{x}\, +\, 2x^2 \sqrt{x}$

. . . . . $x^4\, -\, 16\, =\, (2x^2\, -\, 8)\sqrt{x}$

. . . . . $\frac{x^4\, -\, 16}{2x^2\, -\, 8}\, =\, \sqrt{x}$

Note that x⁴ - 16 = (x² - 4)(x² + 4) and 2x² - 8 = 2(x² - 4).... :wink:

If you get stuck, please reply showing your steps. Thank you!

by **santaclaus** on Tue Mar 10, 2009 1:22 am

what about if as a function f(x) = x² - square root x - 4 + square root x = 0
then, x² - 4=0? will this give me the same answer as squaring both sides, etc.???

by **stapel_eliz** on Tue Mar 10, 2009 2:06 am

Okay, start by adding $\sqrt{x}$ to both sides of your equation. Then solve, keeping in mind that you can only have positive numbers (or zero) inside square roots. :thumb:

by **santaclaus** on Tue Mar 10, 2009 2:14 pm

how or why does the extraneous solution exist? I understand what an extraneous solution is and went and did some research on this, but I would like to know why it exists?

by **stapel_eliz** on Tue Mar 10, 2009 5:49 pm

santaclaus wrote:how or why does the extraneous solution exist?

I don't know how to explain it, other than what it said in the lesson (in the link provided earlier). Sorry! :oops:

For other descriptions, try Google. :wink:

by **santaclaus** on Tue Mar 10, 2009 6:11 pm

Thank you so much for all your help!!!!!! It has really helped alot!!! :clap:

The Purplemath Forums

equation w/square roots: x^2 - square root of x = 4 - sqrt x

equation w/square roots: x^2 - square root of x = 4 - sqrt x

Re: equation w/square roots: x^2 - square root of x = 4 - sqrt x

Re: equation w/square roots: x^2 - square root of x = 4 - sqrt x

Re: equation w/square roots: x^2 - square root of x = 4 - sqrt x

Re: equation w/square roots: x^2 - square root of x = 4 - sqrt x