Question 154120
Remember:
{{{adjacent=a=x}}}
{{{opposite=o=x+1}}}
{{{hypotenuse=h=x+2}}}
consecutive numbers {{{system(x,x+1,x+2)}}}
By Pyth theorem:
{{{h^2=a^2+o^2}}} ------------------> working eqn
{{{(x+2)^2=x^2+(x+1)^2}}}
{{{x^2+4x+4=x^2+x^2+2x+1}}}
{{{2x^2+2x+1-x^2-4x-4=0}}}
{{{x^2-2x-3=0}}}
By Quadratic : {{{a=1}}}, {{{b=-2}}}, & {{{c=-3}}}
{{{x=(-b+-sqrt(b^2-4ac))/2a}}}
{{{x=(-(-2)+-sqrt(-2^2-4*1*-3))/(2*1)}}}
{{{x=(2+-sqrt(4+12))/2}}}
{{{x=(2+-sqrt(16))/2}}}
{{{x=(2+-4)/2}}}
2 VALUES:
{{{x=(2+4)/2=6/2=3}}} -------> USED
{{{x=(2-4)/2=-2/2=-1}}}-------> DON'T USED
So the sides:
{{{adjacent=3}}}
{{{opposite=3+1=4}}}
{{{hypotenuse=3+2=5}}}
In doubt? go back working eqn:
{{{5^2=3^2+4^2}}}
{{{25=9+16}}}
{{{25=25}}}
</pre><font size=4><b>Thank you,
Jojo</pre>