Question

Solve for x: \[\frac{16}{x} - 1 = \frac{15}{x + 1}, x \neq 0, - 1\]

Answer 1

We have been given,\[\frac{16}{x} - 1 = \frac{15}{x + 1}, x \neq 0, - 1\]

Now we solve the above equation as follows,

\[\frac{16 - x}{x} = \frac{15}{x + 1}\]

\[ \Rightarrow (16 - x)(x + 1) = 15x\]

\[ \Rightarrow 16x + 16 - x^2 - x = 15x\]

\[ \Rightarrow 15x + 16 - x^2 - 15x = 0\]

\[ \Rightarrow 16 - x^2 = 0\]

\[ \Rightarrow x^2 - 16 = 0\]

Now we also know that for an equation 

\[a x^2 + bx + c = 0\] the discriminant is given by the following equation:

\[D = b^2 - 4ac\] 

Now, according to the equation given to us, we have, \[a = 1\] ,\[b = 0\] and \[c = - 16\]

Therefore, the discriminant is given as,

\[D = (0 )^2 - 4\left( 1 \right)\left( - 16 \right)\]

\[ = 64\]

Now, the roots of an equation is given by the following equation, \[x = \frac{- b \pm \sqrt{D}}{2a}\]

Therefore, the roots of the equation are given as follows,

\[x = \frac{- 0 \pm \sqrt{64}}{2\left( 1 \right)}\]

\[ = \frac{\pm 8}{2}\]

\[ = \pm 4\]

Therefore, the value of \[x = \pm 4 .\]