Finding solutions to an equation

[Dragonslayer527]

I have the equation 2x + 9Öx -5 = 0

the Ö is the square root sign.

I can't figure out how to get the x from out of the square root to solve for the x.

Any help is appreciated.

 

[stapel]

Just to clarify, I understand your equation to be as follows:

. . . . .2x + 9sqrt(x) - 5 = 0

This is a quadratic in sqrt(x):

. . . . .2[sqrt(x)]² + 9[sqrt(x)] - 5 = 0

Substitute, if this makes things easier:

. . . . .2Y² + 9Y - 5 = 0

Solve by factoring for the values of Y. (You could also use the Quadratic Formula, if you like.) Then back-substitute "sqrt(x)" for "Y", square both sides of your solution equations, and simplify to get your final answers.

Eliz.

 

[soroban]

Hello, Dragonslayer527!

Are you in self-study? .This should have been covered in class.

I have the equation 2x + 9√x - 5 = 0

I can't figure out how to get the x from out of the square root to solve for the x.

Get the radical term alone: . 9√x .= .5 - 2x

Square both sides: . (9√x)² .= .(5 - 2x)²

. . . then we have: . 81x .= .25 - 20x + 4x²

. or the quadratic: . 4x² - 101x + 25 .= .0

Can you finish it now?

 

[tkhunny]

No matter how you find the solution(s), you simply MUST check your answer in the original equation, if there is one. (In a word problem, you have to invent the original equation, so that's a little less conclusive.)

In stapel's version, you will get Y = 1/2 and Y = -5, leading to x = 1/4 and x = sqrt(-5)??? -5 doesn't work. You have to throw it out.

In soroban's version, you will get x = 1/4 and x = 25. 25 doesn't work. You have to throw it out.

There are ways to spot most of these sorts of things, but there really is no substitute for simply CHECKING.