Question

Find the range of values for x for which. (x is postive).

Answer 1

\[\sqrt{x}+\left(\begin{matrix}1 \\ \sqrt{x}\end{matrix}\right)<4\]

Answer 2

sounds fun

Answer 3

not really lol

Answer 4

it really is

Answer 5

i suspect it uses quadratics

Answer 6

yeah because you multiply both sides by \[\sqrt{x}\]

Answer 7

yup

Answer 8

then you let \[a = \sqrt x\]

Answer 9

so \[x = a^2\]

Answer 10

should i help now? or do you have an idea already?

Answer 11

I dunno, what I just did was \[(4\sqrt{x})^2\]

Answer 12

huh? why?

Answer 13

(x+1)^2>16x???

Answer 14

okay first tell me...the questio is \[\sqrt x + \frac{1}{\sqrt x} < 4\]
right?

Answer 15

yes

Answer 16

multiply all terms by \(\sqrt x\) what happens?

Answer 17

\[x+1>4\sqrt{x}\]

Answer 18

why did he sign change? is that a rule in inequality?

Answer 19

the*

Answer 20

typo my bad

Answer 21

ohh

Answer 22

so \[x + 1 < 4\sqrt x\]
i can reqrite this as
\[x - 4\sqrt x + 1 < 0\]
are you following?

Answer 23

yes you just subtracted away the left had side.

Answer 24

right

now llike i said let a = \(\sqrt x\)
so \(a^2 = x\)

the equation becomes
\[\implies a^2 - 4a + 1 < 0\]
do you see how that happened?

Answer 25

how can you just let a= √x

Answer 26

it's just a substitution...to make it easier to see

Answer 27

algebra is just a matter of imagination :)

Answer 28

and also it's a method called "quadratics" which i mentioned earlier

Answer 29

so solve for a.

Answer 30

right

Answer 31

then plug that answer into your A substitution

Answer 32

what ever answer you get it is \(\sqrt x\) so you need to suare the answer to get x

Answer 33

square*

Answer 34

that's a or a^2?

Answer 35

I am not sure I just solve the quadratic.

Answer 36

express your answer as radicals...

Answer 37

\[a = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(1)}}{2(1)}\]
\[\implies \frac{4 \pm \sqrt{16 - 4}}{2}\]
\[\implies \frac{4\pm \sqrt {12}}{2}\]
\[\implies \frac{4 \pm 2\sqrt 3}{2}\]
\[\implies 2 \pm \sqrt 3\]
\[\sqrt x = 2\pm \sqrt 3\]
\[x = (2\pm \sqrt 3)^2\]
\[x = 2^2 \pm 2(2\sqrt 3) + (\sqrt 3)^2\]
\[x = 4 \pm 4\sqrt 3 + 3\]
\[x = 7 \pm 4\sqrt 3\]
so the range is
\[7 - 4\sqrt 3 < x < 7 + 4\sqrt 3\]
does that help?

Answer 38

fun isnt it? :D

Answer 39

you my friend are a genius.

Answer 40

lol. go on and try to review the solution i provided and tell me if you have questions and clarifications

Answer 41

no I am just annoyed I didn't get that myself

Answer 42

Thanks you.

Answer 43

so you understand the process?

Answer 44

yes

Answer 45

wonderful