Me and my GiF

Algebra Level 2

$\large x \lfloor x \rfloor = 28$

Find the positive value of $x$ such that the equation above is satisfied.

Image Credit: Wikimedia Dela Andy Kumahor .

The answer is 5.6.

3 solutions

Led Tasso
Jun 6, 2015

Sorry I'm on my smart phone and it'd take me forever to write a Latex solution.

We have x[x]=28,

x(x-{x})=28, where {x} denotes the fractional part function.

A bit of rearrangement gives {x}=x-28/x.

Now since 0<={x}<1

0<=x-28/x<1

The left part of the inequation yields $\sqrt{28}$ <=x and the right one yields x<(1+ $\sqrt{113}$ )/2

Or approximately 5.29<=x<5.82

Hence [x]=5 and x=28/5.

Abid Junaid
May 24, 2015

IF THE VALUE OF X IS 5, X[X] = 25, AND IF THE VALUE OF X IS 6 X[X] = 36.

THEREFORE THE VALUE OF X LIES BETWEEN 5 & 6.

IF X LIES BETWEEN 5 & 6 THEN [X] = 5

NOW X = 28/[X] = 28/5 = 5.6

Moderator note:

Please try not to use Caps Lock at all times.

Almost perfect. Can you explain why it must be $5<x<6$ simply because you have shown that $x\lfloor x\rfloor = 25$ at $x=5$ and $x\lfloor x\rfloor = 36$ at $x=6$ ?

Bonus question: What would the answer be if I dropped the word "positive" from the question?

Siddharth Singh
May 24, 2015

I tried to take such values of x and [x] so that they may be close to each other,the closest integer values were 4 and 7,but both were integers so I took 5 as the value of [x],then I divided 28 by 5 and got 5.6 which satisfied the equation i.e. 5.6[5.6]=5.6*5=28.

Anyway ,I know this was not an accurate solution but it was what I did:)

Moderator note:

Like the other solution, you did not explain why $x$ must fall between 5 and 6. You should explain that because $f(x) = x \lfloor x \rfloor$ is a strictly non-decreasing function, and because $f(5) < 28 < f(6)$ . Can you finish it off from here?

Me and my GiF

Image Credit: Wikimedia Dela Andy Kumahor .

The answer is 5.6.

3 solutions

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