Roots

Simplify:

3x^2 - 11x - 7 = 0

Finding the roots by plugging in the values a = 3, b = -11, and c = -7 into the quadratic equation \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\), we get:

x = \(\frac { \sqrt{205} + 11}{6}\), x = \(\frac { -\sqrt{205} + 11}{6}\)

We now have that a = \(\frac { \sqrt{205} + 11}{6}\) and b = \(\frac { -\sqrt{205} + 11}{6}\).

Plugging in these values for the expression \(\frac{1}{a+7}+\frac{1}{b+7}\), we get:

\(\frac{1}{\frac { \sqrt{205} + 11}{6}+7}+\frac{1}{\frac { -\sqrt{205} + 11}{6}+7}\)

Simplifying this expression, we get:

\(\frac{1}{\frac { \sqrt{205} + 53}{6}}+\frac{1}{\frac { -\sqrt{205} + 53}{6}}\)

Removing the 6 and multiplying 1 by the reciprocal of \({\frac { \sqrt{205} + 53}{6}}\), we get:

\(\frac{6}{ \sqrt{205} + 53}+\frac{6}{ -\sqrt{205} + 53}\)

We can plug each of these terms in the calculator to get

\(0.089 + 0.155\)

Adding these two values, we get:

\(0.244\)

Answer: 0.244

\($\frac{1}{a+7}+\frac{1}{b+7}.$ \)

Simplify  as

3x^2 -11x -7  =  0

We have the form mx^2 + nx + c  = 0

m =3 , n = -11   c = -7

Product of  roots   = ab = c/m   =  -7/3

Sum of  roots =  

a + b = -n/m  =   11/3          

1/ (a + 7)  +  1 /(b + 7)   =

[ a + 7 + b + 7 ] / [ (a + 7) (b + 7) ]  =

[ a + b + 14 ] / [ ab + 7 (a + b) + 49 ]  = 

[ 11/3 + 14 ]  / [ -7/3 + 7(11/3) + 49 ]  =

[ 53/3 ] / [ 217 /3 ]  =

53 / 217