Finding m given that&nbsp;(m2−1)x2−3(3m−1)x+18=0(a) Suppose m is an integer so that&nbsp;(m2−1)x2−3(3m−1)x+18=0&nbsp;has two positive integer roots. Find m.(b) Now, suppose that we have a triangle&nbsp;ABC&nbsp;with sides&nbsp;a,b,c&nbsp;such that&nbsp;c=23m2+a2m−8a=0m2+b2m−8b=0.Find the area of ABC.

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Finding m given that&amp;nbsp;(m2−1)x2−3(3m−1)x+18=0(a) Suppose m is an integer so that&amp;nbsp;(m2−1)x2−3(3m−1)x+18=0&amp;nbsp;has two positive integer roots. Find m.(b) Now, suppose that we have a triangle&amp;nbsp;ABC&amp;nbsp;with sides&amp;nbsp;a,b,c&amp;nbsp;such that&amp;nbsp;c=23m2+a2m−8a=0m2+b2m−8b=0.Find the area of ABC.

WigwrannyErarmbmk · Accepted Answer

a) (m2−1)x2−3(3m−1)x+18=((m−1)x−3)((m+1)x−6),which since m2−1≠0, givesx=3m−1or x=6m+1and easy to see that only m=2 is valid.

cutimnm135imsa · Accepted Answer

b) Since a2−4a+2=0 and b2−4b+2=0, we obtain following triples for sides-lengths of the triangle:(2+2,2+2,23)and (2+2,2−2,23).Now, for calculating of the area we can use the following formula:SΔABC=142(a2b2+a2c2+b2c2)−a4−b4−c4.For the first we obtain:S=9+122.For the second we obtain S=1.

Finding m given that \(\displaystyle{\left({m}^{{2}}-{1}\right)}{x}^{{2}}-{3}{\left({3}{m}-{1}\right)}{x}+{18}={0}\) (a) Suppose m

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