GATE Electrical 2018 | GA | Question: 4

Question

For what values of $k$ given below is  $\dfrac{(k + 2)^2}{(k - 3)}$ an integer?

• A. $4 , 8 , 18 $   
• B. $4 , 10 , 16$
• C. $ 4 , 8 , 28 $          
• D. $8 , 26 , 28$

Answer 1

We'll try to eliminate the wrong options

• when $k = 4,   \dfrac{(4 + 2)^2}{4 - 3} = 36$ (integer)
• when $k = 8,  \dfrac{(8 + 2)^2}{8 - 3}$ = $\dfrac{100}{5} = 20$ (integer)
• when $k = 10, \dfrac{(10 + 2)^2}{10 - 3}$ = $\dfrac{144}{7} = 20.57$ (not an integer)
• when $k = 16, \dfrac{(16 + 2)^2}{16 - 3}$ = $\dfrac{324}{13} = 24.92$ (not an integer)
• when $k = 18, \dfrac{(18 + 2)^2}{18 - 3}$ = $\dfrac{400}{15} = 26.66$ (not an integer)
• when $k = 26, \dfrac{(26 + 2)^2}{26 - 3}$ = $\dfrac{784}{23} = 34.08$ (not an integer)
• when $k = 28, \dfrac{(28 + 2)^2}{28 - 3}$ = $\dfrac{900}{25} = 36$ (integer)

Correct option  (C) 4, 8, 28

Answer 2

Answer can be found out using synthetic division.

 

Comments

nice approach @Veenit

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explaination of this solution:

$\dfrac{(k+2)^2}{(k-3)} = \dfrac{(k^2 + 4k+4)}{(k-3)} = (k+7) + \dfrac{25}{(k-3)}$

now, how I have arrived to the last form ?

1) first method is to divide $(k^2 + 4k+4)$ by $(k-3)$. so, you will get $(k+7)$ as quotient and $25$ as remainder like this:

2) second method is like this:

we want to convert this: $(k^2 + 4k+4)$, such that one of it's factor remains $k=3$. so, that $(k-3)$ can divide it.

so, basically the intended format is like this: $(k^2 + 4k+4) = (k-3)(k- \alpha) + C$  ($\alpha$ is the second root and $C$ is the constant term)

$(k^2 + 4k+4) = (k-3)(k- \alpha) + C$

$(k^2 + 4k+4) = k^2 - \alpha k -3k +3 \alpha + C$

$(k^2 + 4k+4) = k^2 - (\alpha + 3)k +3 \alpha + C$

comparing both sides,

a) $-(\alpha + 3) = 4 \quad \Rightarrow \quad \alpha + 3 = -4 \quad \Rightarrow \quad \boxed{\alpha = -7}$

b) $3 \alpha + C = 4 \quad \Rightarrow \quad 3 \times -7 +C = 4 \quad \Rightarrow \quad -21 + C = 4 \quad \Rightarrow \quad \boxed{C = 25}$

hence, $(k^2 + 4k+4) = (k-3)(k- (-7)) + 25 = \boxed{(k-3)(k+7) + 25}$

3) third method is like this:

we have this equation: $(k^2 + 4k+4)$ and basically we want:

$(k^2 + 4k+4) \quad \Rightarrow \quad ((k^2 + 4k + B) + 4 -B) \quad \Rightarrow \quad (k-3)(k- \alpha) + C$

so, we have $((k^2 + 4k + B) + 4 -B)$, where $B$ is a constant.

now, we want that $(k^2 + 4k + B)$ should have a root as $k=3$.

sum of roots: $\alpha + \beta = \dfrac{-4}{1} \quad \Rightarrow \quad \alpha + 3 = -4 \quad \Rightarrow \quad \boxed{\alpha = -7}$

product of roots: $\alpha \cdot \beta = \dfrac{B}{1} \quad \Rightarrow \quad -7 \times 3 = B \quad \Rightarrow \quad \boxed{B = -21}$

so, $(k^2 + 4k + B) = (k^2 + 4k -21)$

now, $(k^2 + 4k+4)$
$= (k^2 + 4k + B) + 4 -B $
$= (k^2 + 4k - 21) + 4 -(-21)$
$= (k^2 + 7k - 3k - 21) + 4 + 21$
$= (k-3)(k+7) + 25$

hence, $\boxed{(k^2 + 4k+4) = (k-3)(k+7) + 25}$

Now, as we got this:

$\dfrac{(k+2)^2}{(k-3)} = (k+7) + \dfrac{25}{(k-3)}$

if $k$ itself is an integer then $(k+7)$ will also be integer but for this $\dfrac{25}{(k-3)}$ to be integer, $(k-3)$ needs to be a factor of $25$.

factors of $25 = 1, 5, 25$

so, $k -3 = 1, 5, 25 \quad \Rightarrow \quad \boxed{k = 4, 8, 28}$

Answer 3

We'll try and eliminate the wrong options

when K = 4  ==>   $\dfrac{(4 + 2)^2}{(4 - 3)}$ = 36 (integer)

when K = 8  ==>   $\dfrac{(8 + 2)^2}{(8 - 3)}$ = $\dfrac{(100)}{(5)}$ = 20 (integer)

when K = 10  ==>   $\dfrac{(10 + 2)^2}{(10 - 3)}$ = $\dfrac{(144)}{(7)}$ = 20.57 (not an integer)

when K = 16  ==>   $\dfrac{(16 + 2)^2}{(16 - 3)}$ = $\dfrac{(324)}{(13)}$ = 24.92 (not an integer)

when K = 18  ==>   $\dfrac{(18 + 2)^2}{(18 - 3)}$ = $\dfrac{(400)}{(15)}$ = 26.66 (not an integer)

when K = 26  ==>   $\dfrac{(26 + 2)^2}{(26 - 3)}$ = $\dfrac{(784)}{(23)}$ = 34.08 (not an integer)

when K = 28  ==>   $\dfrac{(28 + 2)^2}{(28 - 3)}$ = $\dfrac{(900)}{(25)}$ = 36 (integer)

By Verifying all the options we found out the correct option as C) 4, 8, 28