A quadratic equation given roots is solved by using the relation
from the question,
sum of the root will be
and also, product of the roots will
be
now substituting the sum of roots and product of roots into the equation
as the quadratic equation for the root 3/4 and -4
We are given sides of the triangle by expression
First side : n
Second side : m = 5n-5.
Third side : l = 5/4 m - n.
Perimeter is 90.
Therefore, first equation could be set :
n +m +l = 90 --------------equation(1)
Second equation would be
m = 5n-5 --------------equation(2)
Third equation would be
l = 5/4 m - n --------------equation(3)
Let us solve the system of equations by substitution.
Substituting m = 5n-5 and l = 5/4 m - n in first equation.
We get
n +5n-5 +5/4 m - n= 90
Now, substituting m = 5n-5 in above equation we get
n +5n-5 +5/4 (5n-5) - n= 90
5n + 25n/4 -5 - 25/4 =90
5n + 6.25n -5-6.25 = 90.
11.25n-11.25 = 90
Adding 11.25 on both sides, we get
11.25n-11.25+11.25 = 90+11.25
11.25n = 101.25.
Dividing both sides by 11.25, we get
n=9.
Plugging n=9 in 2nd equation.
m = 5(9)-5 = 45 -5 = 40.
Plugging n=9 and m=40 in first equation, we get
9 +40 +l = 90
49 +l = 90.
Subtracting 49 from both sides, we get
49-49 +l = 90-49
l = 41.
Therefore, sides are of lengths l = 41, m= 40 and n=9.
