Question

If \(a\) and \(b\)are positive numbers, prove that the equation

\(\frac{a}{{{x^3} + 2{x^2} - 1}} + \frac{b}{{{x^3} + x - 2}} = 0\)

has at least one solution in the interval \(\left( { - 1,1} \right)\).

Short answer

It is proved that the given equation has at least one solution in the interval \(\left( { - 1,1} \right)\).

Step-by-step solution

Intermediate value theorem

Consider thefunction\(f\)iscontinuous on the closed interval\(\left( {a,b} \right)\)and let\(N\)be any number between\(f\left( a \right)\)and\(f\left( b \right)\)with\(f\left( a \right) \ne f\left( b \right)\). Then there exists anumber\(c\)in \(\left( {a,b} \right)\) such that \(f\left( c \right) = N\).

Step 2:Prove that the given equation has at least one solution

The equation is as shown below:

\(\begin{array}{c}\frac{a}{{{x^3} + 2{x^2} - 1}} + \frac{b}{{{x^3} + x - 2}} = 0\\a\left( {{x^3} + x - 2} \right) + b\left( {{x^3} + 2{x^2} - 1} \right) = 0\end{array}\)

Consider\(p\left( x \right)\)as the left-hand side of this equation. According to the Intermediate Value theorem,\(p\left( { - 1} \right) = - 4a < 0\), and\(p\left( 1 \right) = 2b > 0\)because the function\(p\)is continuous on the closed interval\(\left( { - 1,1} \right)\)there is a number\(c\)in\(\left( { - 1,1} \right)\)such that

\(p\left( c \right) = 0\).

It is observed that the only solution of either denominator that is in the interval \(\left( { - 1,1} \right)\)is\(\frac{{\left( { - 1 + \sqrt 5 } \right)}}{2} = r\), however\(p\left( r \right) = \frac{{\left( {3\sqrt 5 - 9} \right)a}}{2} \ne 0\).

Hence, the number\(c\)is not a solution of either denominator, therefore,\(p\left( c \right) = 0\). This gives that\(x = c\)is a solution of the equation.

Thus, it is proved that the given equation has at least one solution.