solution of √ (x + 7) < x

[sanju09]

What is the solution of √ (x + 7) < x?
(A) x > 2
(B) x > ½ √30
(C) x > ½ (1 + √29)
(D) x > 1 + ½ √29
(E) x > 1 + ½ √30


[spoiler]Made up[/spoiler]

[shovan85]

IMO C

Square both sides and get the eqn of the form x^2 - x - 7 > 0.

Solving for this we can get either

x> [1+sqrt(29)]/2 or x> [1-sqrt(29)]/2

[neerajkumar1_1]

root(x+7)<x

x will have to be +ve since root(x+7) will always be +ve...

now squaring both sides we get

x+7 < x^2
=> x^2 - x - 7 > 0
solve for roots of x using formula [-b +- root(b^2 - 4ac)]/2a

we get x to be 1/2(1 + root(29)) and 1/2(1 - root(29))

so
(x - 1/2(1 + root(29)))(x- 1/2(1 - root(29))) > 0

hence x > 1/2(1 + root(29))
or x> 1/2(1 - root(29))

first one is among the answer choices...

PIck C

[diebeatsthegmat]

What is the solution of √ (x + 7) < x?
(A) x > 2
(B) x > ½ √30
(C) x > ½ (1 + √29)
(D) x > 1 + ½ √29
(E) x > 1 + ½ √30


[spoiler]Made up[/spoiler]

only C.,..
hey, where is the source? i saw you posted so many nice maths but some Ps is weird...

[pzazz12]

root(x+7)<x

x will have to be +ve since root(x+7) will always be +ve...

now squaring both sides we get

x+7 < x^2
=> x^2 - x - 7 > 0
solve for roots of x using formula [-b +- root(b^2 - 4ac)]/2a

we get x to be 1/2(1 + root(29)) and 1/2(1 - root(29))

so
(x - 1/2(1 + root(29)))(x- 1/2(1 - root(29))) > 0

hence x > 1/2(1 + root(29))
or x> 1/2(1 - root(29))

first one is among the answer choices...

PIck C

thank u.......is there any other way to solve these type of problems....

[Rahul@gurome]

We get that x^2 - x - 7 > 0.
Instead of using the formula for roots we can write that x^2 - x =(x - ½)^2 - ¼.
Or x^2 - x - 7 > 0 can be written as (x - ½)^2 - ¼ -7 > 0.
Or we have that (x - ½)^2 > 29/4.
So either (x-1/2) > sqrt(29)/2 or (x-1/2) < -sqrt(29)/2.
So either x >1/2(1 + sqrt29) or x < ½(1 - sqrt29) {this is negative}.Now x cannot be negative because sqrt(x+7) is positive and positive can never be less than negative.
So the only possible range for x is x >1/2(1 + sqrt29).