Question 156174
let {{{x=adjacent}}}
{{{x+2=opposite}}}, 2 meters longer
{{{10m=hypotenuse}}}
by Pyth theorem:
{{{hyp^2=adj^2+opp^2}}}
{{{10^2=x^2+(x+2)^2}}}
{{{100=x^2+x^2+4x+4}}}
{{{2x^2+4x+4-100=0}}} ---> {{{2x^2+4x-96=0}}}, Divide the whole eqn by 2 to reduce
{{{cross(2)x^2/cross(2)+cross(4)2x/cross(2)-cross(96)48/cross(2)=0}}}
{{{x^2+2x-48=0}}}
BY QUADRATIC, {{{a=1}}}, {{{b=2}}}, & {{{c=-48}}}
{{{x=(-b+-sqrt(b^2-4ac))/(2a)}}}
{{{x=(-2+-sqrt(2^2-4*1*-48))/(2*1)}}}
{{{x=(-2+-sqrt(4+192))/2=(-2+-sqrt(196))/2}}}
{{{x=(-2+-14)/2}}}, 2 values:
{{{x=(-2+14)/2=12/2=6}}} ---------------------> TO BE USED
{{{x=(-2-14)/2=-16/2=-8}}} -------------------> DON'T USE (-)
So, {{{adjacent=x=6meters}}}
{{{opposite=x+2=6+2=8meters}}}
In doubt? go back Pyth Theorem:
{{{10^2=6^2+8^2}}}
{{{100=36+64}}}
{{{100m=100m}}}
Thank you,
Jojo