Question 700717
{{{((3x^2+13x+4)/(x^2-4)) /( (4x+16)/(x+2))= ((3x^2+13x+4)/(x-2)) *(( 1/4)x+16)}}}



factor: {{{3x^2+13x+4=(x+4)(3x+1)}}} and {{{x^2-4=(x-2)(x+2)}}}



{{{(((x+4) (3x+1))/((x-2)(x+2)))/(4(x+4)/(x+2))= (((x+4) (3x+1))/(x-2)) *(( 1/4)x+16)) }}}



{{{((cross((x+4)) (3x+1))/((x-2)cross((x+2))))/(4cross((x+4))/cross((x+2)))= (((x+4)(3x+1))/(x-2)) *(( x/4+16)) }}}


{{{(3x+1)/4(x-2)= (((x+4)(3x+1))/(x-2)) *(( x+64)/4) }}}


{{{(3x+1)/4(x-2)= ((x+4)(3x+1)( x+64))/4(x-2) }}}....if denominators same, nominators are same too, so, we can write

{{{3x+1= (x+4)(3x+1)( x+64) }}}

{{{0= (x+4)(3x+1)( x+64)-(3x+1) }}}

{{{(3x+1)((x+4)( x+64)-1)}}}

{{{(3x+1)(x^2+64x+4x+256-1)=0}}}

{{{(3x+1)(x^2+68x+255)=0}}}

solutions:

if {{{3x+1=0}}}.....=>.....{{{3x=-1}}}...=>.....{{{highlight(x=-1/3)}}}



{{{x^2+68x+255=0}}}..use quadratic formula


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}


{{{x = (-68 +- sqrt( 68^2-4*1*255 ))/(2*1) }}}


{{{x = (-68 +- sqrt(4624-1020 ))/2 }}}


{{{x = (-68 +- sqrt(3604 ))/2 }}}


{{{x = (-68 +- sqrt(4*901 ))/2 }}}


{{{x = (-68 +- 2sqrt(901 ))/2 }}}


{{{x = (-cross(68)34 +- cross(2)sqrt(901 ))/cross(2) }}}


{{{x = (-34 +- sqrt(901 )) }}}

so, {{{x = (-34 + sqrt(901 )) }}} or {{{x = (-34 - sqrt(901 )) }}}


since {{{sqrt(901)}}}≈ {{{30}}}, we have

{{{x = -34 + 3 }}} 

{{{highlight(x = -31) }}} 

or 

{{{x = -34 -30 }}}

{{{highlight(x = -64)  }}}



simplify 
{{{a^(1/2) * b^3/4}}}...if you have this, than

{{{sqrt(a) *( b^3/4)}}}

but, if you have {{{a^(1/2) * b^(3/4)}}}, than

{{{sqrt(a) *root(4, b^3)}}}



Simplify 

{{{n*sqrt(a)=a^x}}}...simplest form


Simplfy 

{{{8^2x+3=2^a * b^x}}}.......I am not quite sure do you have {{{8^2x}}} or {{{8^(2x)}}}


Simplify {{{3x*2^3x+2}}}......same here,do you have {{{2^3x}}} or {{{2^(3x)}}}