Question

factorise ( x^2+1/x^2)-4(x+1/x)+6

factorise2x^2+2root6xy+3y^2

Answer 1

Hi ...dear..
here is your answer...
.
..I added +2and. -2 in second step to complete the square...!!
and after that let x+1/x be y ...so that it will become simple quadratic polynomial...!!!!.
and ans will (x+1/x -2^2....
......and second question is a a whole square .....hope it helped you!!!.
Regards
Brainly Star Community.
#shubhendu

Answer 2

Answer:

$2x^{2} +2\sqrt{6}xy+3y^{2}==(\sqrt{2}x+\sqrt{3}y )^{2}$

$(x^{2} +\frac{1}{x^{2} } )-4(x+\frac{1}{x} )+6=(x+\frac{1}{x}-2 )^{2}$

Step-by-step explanation:

Given:  $(x^{2} +\frac{1}{x^{2} } )-4(x+\frac{1}{x} )+6$ and $2x^{2} +2\sqrt{6}xy+3y^{2}$

To factorize the given expression.

For $(x^{2} +\frac{1}{x^{2} } )-4(x+\frac{1}{x} )+6$.

$(x^{2} +\frac{1}{x^{2} } )-4(x+\frac{1}{x} )+6=(x^{2} +\frac{1}{x^{2} }+2-2 )-4(x+\frac{1}{x} )+6$

                                       $=(x+\frac{1}{x} )^{2} -2-4(x+\frac{1}{x} )+6$

Let $x+\frac{1}{x}=t$

So, $(x+\frac{1}{x} )^{2} -2-4(x+\frac{1}{x} )+6=t^{2} -4t+4$

                                                 $=(t-2)^{2}$

Back substituting we get,

$(t-2)^{2} =(x+\frac{1}{x}-2 )^{2}$

So, $(x^{2} +\frac{1}{x^{2} } )-4(x+\frac{1}{x} )+6=(x+\frac{1}{x}-2 )^{2}$

For $2x^{2} +2\sqrt{6}xy+3y^{2}$

Using identity, $(a+b)^{2} =a^{2}+2ab+b^{2}$

Here, $2x^{2} +2\sqrt{6}xy+3y^{2}=(\sqrt{2}x)^{2} +2(\sqrt{2}x)(\sqrt{3} y)+(\sqrt{3}y )^{2}$

                                        $=(\sqrt{2}x+\sqrt{3}y )^{2}$

#SPJ2