Question 8615
Jennifer's solution is NOT correct!  When you add the base and to the height you do NOT get the hypotenuse, as she suggested!!


Area of a triangle: 
{{{A = (1/2)* (bh)}}}

{{{ (1/2)*(4x-4)*(x-3)= (4x^2+10)^2}}}

{{{ (2x-2)*(x-3)= (4x^2+10)^2}}}

{{{2x^2 -6x-2x + 6 = 16x^4 + 80x^2 + 100 }}}
{{{0 = 16x^4 + 78x^2 + 8x + 94 }}}


Since this seems unreasonable to solve, perhaps you stated the problem wrong!!  Recheck the problem.  I was thinking that maybe the area is just {{{4x^2 + 10}}}, without the "squared."


{{{ (2x-2)*(x-3)= (4x^2+10) }}}

{{{2x^2 -6x-2x + 6 = 4x^2 + 10 }}}
{{{0 = 2x^2 + 8x + 4 }}}
{{{0 = 2(x^2 + 4x + 2) }}}

Solve by quadratic formula, and it just comes out ugly??  Recheck your copying of the problem to make sure that is correct.  
 

R^2 at SCC