Quadrilateral ABCD is inscribed in this circle.
1 answer:
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Let
x---------> first positive integer
x+1------> second positive integer
x+2-----> third positive integer
we know that
(x+1)*(x+2)=72-------> x² +2x+x+2=72 -------> x² +3x-70=0
using a graph tool-------> <span>I solve the quadratic equation
</span>see the attached figure
the roots are
x1=-10
x2=7
the answer is
first positive integer is x=7
second positive integer is x+1=8
third positive integer is x+2=9
x---------> first positive integer
x+1------> second positive integer
x+2-----> third positive integer
we know that
(x+1)*(x+2)=72-------> x² +2x+x+2=72 -------> x² +3x-70=0
using a graph tool-------> <span>I solve the quadratic equation
</span>see the attached figure
the roots are
x1=-10
x2=7
the answer is
first positive integer is x=7
second positive integer is x+1=8
third positive integer is x+2=9