Quadratic equation: 3x + 10 sqrt[x] - 8 = 0

[jshaziza]

3x+10(x)-8=0

(x) is my symbol for the square root of x. Don't know how to put the real symbol in.

This is what I did to solve this problem:
u=(x) and u^2=x
u^2+10u-8=0
a=1, b=10, c=-8
I then used the quadratic formula and got u= -5+(33) and -5-(33).
Next I plugged u back in for x.
(x)^2=(-5+/-(33)^2
x=58 and x=-8

When I put these numbers into the original formula though it didn't work, so somewhere along the line I did something wrong. If some could point it out, that would be great. thx.[/u]

 

[skeeter]

3x + 10sqrt(x) - 8 = 0 ... domain is x > 0

3u² + 10u - 8 = 0

(3u - 2)(u + 4) = 0

u = 2/3 ... sqrt(x) = (2/3), x = 4/9

u = -4 ... sqrt(x) = -4 ... x has no solution.

 

[galactus]

Re: Quadratic equation

3x+10(x)-8=0

(x) is my symbol for the square root of x. Don't know how to put the real symbol in.

Why not just type '10sqrt(x)'. Better yet, \(\displaystyle 10\sqrt{x}\).

This is what I did to solve this problem:
u=(x) and u^2=x
3u^2+10u-8=0
a=3, b=10, c=-8
I then used the quadratic formula and got u=2/3 and -4.
Next I plugged u back in for x.
(x)^2=(-5+/-(33)^2
x=58 and x=-8

You made a good substitution. \(\displaystyle u=\sqrt{x}, \;\ u^{2}=x\)

That gives \(\displaystyle \L\\3u^{2}+10x-8=0\)

Now, solve that quadratic and remember that \(\displaystyle u=\sqrt{x}\)

 

[jshaziza]

Skeeter why did you say x has no solution? When I plugged in x=4/9 it worked out.

 

[galactus]

-4 has no solutions. 4/9 is the only good solution from the two