Chapter 7.8, Problem 26E

To determine

The value of t from the given equation.

Answer to Problem 26E

The value of variable by factorization Method or using Quadratic formula is $t=0,8,-9$ .

Explanation of Solution

Given information: - The given factoring equation are

  $2t^5-2t^4-144t^3=0$

Calculation: - According to given equation

  $2t^5-2t^4-144t^3=0$

Take 2 $t^3$ common in this equation

  $2t^3(t^2+t-72)=0$ ----------------------------(1)

Then,

, $\begin{array}{l}2t^3=0\\ t^3=0\\ t=\sqrt[3]{0},t=0\end{array}$

[Any root of zero is zero]

  $\begin{array}{l}{t}^2+t-72=0\\ t^2+9t-8t-72=0\\ t(t+9)-8(t+9)=0\\ t+9=0,t-8=0\\ t=-9,t=8\end{array}$

Thus, the value of variable of k is $=0,8,-9$ .

By Quadratic equation- According to the Quadratic Equation formula,

  $A{n}^2+Bn+C=0$

Where A, B & C are numbers called coefficient

  $n=\frac{-B\pm \sqrt{B^2-4AC}}{2A}$

In given case,

  $\begin{array}{l}A=1\\ B=1\\ C=-72\end{array}$

Accordingly

  $\begin{array}{l}{B}^2-4AC\\ {(1)}^2-4\times 1\times -72\\ 1+288=0\\ 289=0\\ B^2-4AC=289\end{array}$

By applying the Quadratic formula

  $n=\frac{-(1)\pm \sqrt{289}}{2\times 1}$

  $n=\frac{1\pm 17}{2}$

  $\begin{array}{l}n=\frac{-1+17}{2},n=\frac{-1-17}{2}\\ n=\frac{16}{2},n=\frac{-18}{2}\\ n=8,-9\end{array}$

  $2t^3=0$ [from equation (1)]

  $\begin{array}{l}{t}^3=0\\ t=\sqrt[3]{0}\\ t=0\end{array}$

Thus, the value of variable t is $=0,8,-9$ .