Question

Solve 2x+9(x^1/2)-5 Please show steps!

Answer 1

\[= 2x + 9x ^{1.5} - 5\]
ok?

Answer 2

@Kenneditodd what does the question want?
the mesure of x?

Answer 3

That means the square root of x....the equation is equal to zero, and yes it wants x.....I know the answer I just need to know how to get it

Answer 4

\[x^{1.5} = \sqrt{x ^{3}}\]
ok?

Answer 5

@Kenneditodd , OK?

Answer 6

No it's x=1/4

Answer 7

@PFEH.1999

Answer 8

ok ok

Answer 9

but I didn't find the answer...

Answer 10

Yes....and I think you misunderstood the problem, it's 2x+9 times the square root of x -5 @PFEH.1999

Answer 11

X-5 is not a binomial btw

Answer 12

@sweetsandeverything yes

Answer 13

If we are looking for the zero of the expression, we can make a substitution to make it easier. Let u = sqrt x, then\[0=u^2+9u-5=(2u-1)(u+5) \implies u=\frac{1}{2}= \sqrt {x} ~or~u=-5\]\[\implies x=\frac{1}{4}\]The solution u=-5 is discarded, since the square root is positive.

Answer 14

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

Let \(u = \sqrt x \), then \(u^2 = x\)

Substitute in original eqution:

\(2u^2 + 9u - 5 =0 \)

\((2u - 1)(u + 5) =0 \)

\(2u - 1 = 0\) or \(u + 5 = 0\)

\(2u = 1\) or \(u = -5\)

\(u = \dfrac{1}{2} \) or \(u = -5\)

Since x = u^2,

\(x = \left( \dfrac{1}{2} \right)^2 \)

\(x = \dfrac{1}{4} \)

or

\(x = (-5)^2 \)

\(x = 25\)

When you check the two solutions in the original eqaution, x = 25 does not work, so the solutioon is

\(x = \dfrac{1}{4} \)