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For a period of d days an account balance can be modeled by f(d) = d^ 3 -2d^2 -8d +3 when was the balance $38

User Ali Azhar
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1 Answer

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Given a modelled account balance for the period of d days as shown below:


\begin{gathered} f(d)=d^3-2d^2-8d+3 \\ \text{where,} \\ f(d)\text{ is the account balance} \\ d\text{ is the number of days} \end{gathered}

Given that the account balance is $38, we would calculate the number of days by substituting for f(d) = 38 in the modelled equation as shown below:


\begin{gathered} 38=d^3-2d^2-8d+3 \\ d^3-2d^2-8d+3-38=0 \\ d^3-2d^2-8d-35=0 \end{gathered}

Since all coefficients of the variable d from degree 3 to 1 are integers, we would apply apply the Rational Zeros Theorem.

The trailing coefficient (coefficient of the constant term) is −35.

Find its factors (with plus and minus): ±1,±5,±7,±35. These are the possible values for dthat would give the zeros of the equation

Lets input x= 5


\begin{gathered} 5^3-2(5)^2-8(5)-35=0 \\ 125-2(25)-40-35=0 \\ 125-50-75=0 \\ 125-125=0 \\ 0=0 \end{gathered}

Since, x= 5 is a zero, then x-5 is a factor.


\begin{gathered} d^3-2d^2-8d-35=(d-5)(d^2+3d+7)=0 \\ (d-5)(d^2+3d+7)=0 \\ d-5=0,d^2+3d+7=0 \\ d=0, \end{gathered}
\begin{gathered} \text{simplifying } \\ d^2+3d+7\text{ would give} \\ d=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ a=1,b=3,c=7 \end{gathered}
\begin{gathered} d=\frac{-3\pm\sqrt[]{3^2-4*1*7}}{2*1} \\ d=\frac{-3\pm\sqrt[]{9-28}}{2} \\ d=\frac{-3\pm\sqrt[]{-17}}{2} \end{gathered}

It can be observed that the roots of the equation would give one real root and two complex roots

Therefore,


d=5,d=\frac{-3\pm\sqrt[]{-17}}{2}

Since number of days cannot a complex number, hence, the number of days that would give a balance of $38 is 5 days

User Bart Wegrzyn
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