Asked by Grace
Solve:
((10)/(x^2-3x+45))=x+1
((10)/(x^2-3x+45))=x+1
Answers
Answered by
Damon
(x+1) (x^2-3x+45) = 10
x^3 - 2 x^2 + 42 x + 45 = 10
x^3 - 2 x^2 + 42 x + 35 = 0
search for a root
x , function
1 , 76
0 , 35
-1 , -10
so at least one root between -1 and 0
-.5 , 13.375
-.75 , 1.95
-.80 , -.392
That is pretty close, we could keep going of course but then
divide
x^3 - 2 x^2 + 42 x + 35
by
(x+0.8)
and get a quadratic which you can solve with the quadratic equation
x^3 - 2 x^2 + 42 x + 45 = 10
x^3 - 2 x^2 + 42 x + 35 = 0
search for a root
x , function
1 , 76
0 , 35
-1 , -10
so at least one root between -1 and 0
-.5 , 13.375
-.75 , 1.95
-.80 , -.392
That is pretty close, we could keep going of course but then
divide
x^3 - 2 x^2 + 42 x + 35
by
(x+0.8)
and get a quadratic which you can solve with the quadratic equation
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