Answer: The required positive solution is x = 1.3.
Step-by-step explanation: We are given to find the positive solution of the following quadratic equation:
[tex]2x^2+3x-8=0~~~~~~~~~~~~~~~~~~(i)[/tex]
The solution set of the quadratic equation [tex]ax^2+bx+c=0,~~a\neq 0[/tex] is given by
[tex]x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.[/tex]
From equation (i), we have
a = 2, b = 3 and c = -8.
Therefore, the solution set of equation (i) is given by
[tex]x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\\\\Rightarrow x=\dfrac{-3\pm\sqrt{3^2-4\times 2\times (-8)}}{2\times 2}\\\\\\\Rightarrow x=\dfrac{-3\pm\sqrt{9+64}}{4}\\\\\\\Rightarrow x=\dfrac{-3\pm\sqrt{67}}{4}\\\\\Rightarrow x=\dfrac{-3+\sqrt{67}}{4},~~~x=\dfrac{-3-\sqrt{67}}{4}\\\\\\\Rightarrow x=\dfrac{-3+8.18}{4},~~\Rightarrow x=\dfrac{-3-8.18}{4}\\\\\\\Rightarrow x=\dfrac{5.18}{4},~~\Rightarrow x=-\dfrac{11.8}{4}\\\\\\\Rightarrow x=1.295,~~~~~\Rightarrow x=-2.95.[/tex]
So, the required positive solution is x = 1.295.
Rounding to nearest hundredth, we get x = 1.3.
Thus, the required positive solution is x = 1.3.
