Question

find if x+ 1/x=3 is a perfect square or not and please tell the process​

Answer 1

(X+1)(X+2)(X+3)(X+k)+1=(x^2+3x+2) × (x^2+ 3x+kx+3k) +1=x^4+3x^3+kx^3 +3kx^2 +3x^3+9x^2+3kx^2+9kx +2x^2+6x+2kx+6k +1 =x^4+x^3(3+k+3)+x^2(3k+9+3k+2)+x(9k+2k+6)+(6k+1)= x^4+(k+6)x^3+(6k+11)x^2+(11k+6)x+(6k+1)=(x^2+ax+b)^2 say , where 2a=k+6, a^2+2b=6k+11, 2ab=11k+6, b^2=6k+1 ,

Solving for integral values gives us k = 4, b = 5, a = 5 as foollows.

2ab = 11k + 6, 2a = k + 6 gives us

b = 11k + 6 / k + 6 &

b^2 = 6k + 1 gives us eqn k^2 —8k + 16 = 0

So k = 4

With this value of k , the given expression is a perfect square for any value of x

Now Our expression is (x+1)(x+2)(x+3)(x+4)+1 & we claim that this is perfect square for any value of x with the square root as, x^2+5x+5

Answer 2

Step-by-step explanation:

${\red{\bf{GIVEN:-}}} \\ {\gray{\bf{x - \frac{1}{2} = 5}}} \\ \\ {\blue{\bf{so \: squaring \: both \: sides \: : ({x - \frac{1}{2}) = {25} }}}} \\ {\pink{\bf{ {x}^{2} + \ { \frac{1}{2} }^{2} - 2x \times \frac{1}{x} = 25 }}} \\ {\green{\bf{ {x}^{2} + \frac{1}{ {x}^{2} } = 25 + 2}}}$

${\purple{\bf{ {x}^{2} + \frac{1}{ {x}^{2} = 27 } }}}$

HOPE IT HELPS ✨✨