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

Simplificatation, evaluation, linear equations, linear graphs, linear inequalities, basic word problems, etc.
santaclaus
Posts: 41
Joined: Tue Mar 03, 2009 12:17 am
Contact:

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

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.

stapel_eliz
Posts: 1687
Joined: Mon Dec 08, 2008 4:22 pm
Contact:
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.

santaclaus
Posts: 41
Joined: Tue Mar 03, 2009 12:17 am
Contact:

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

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?

stapel_eliz
Posts: 1687
Joined: Mon Dec 08, 2008 4:22 pm
Contact:
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

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

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)....

santaclaus
Posts: 41
Joined: Tue Mar 03, 2009 12:17 am
Contact:

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

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.???

stapel_eliz
Posts: 1687
Joined: Mon Dec 08, 2008 4:22 pm
Contact:
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.

santaclaus
Posts: 41
Joined: Tue Mar 03, 2009 12:17 am
Contact:

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

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?

stapel_eliz
Posts: 1687
Joined: Mon Dec 08, 2008 4:22 pm
Contact:
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!