Respuesta :
Given the equation:
[tex]2x^2+7x-15=0[/tex]Where r and s are two solutions of the equation and r is greater than x.
Let's solve for r - s.
To find the solutions, let's use the quadratic formula:
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]Where:
a = 2
b = 7
c = -15
Thus, we have:
[tex]\begin{gathered} x=\frac{-7\pm\sqrt[]{7^2-4(2)(-15)}}{2(2)} \\ \\ x=\frac{-7\pm\sqrt[]{49+120}}{4} \\ \\ x=\frac{-7\pm\sqrt[]{169}}{4} \\ \\ x=\frac{-7\pm13}{4} \\ \end{gathered}[/tex]Solving further:
[tex]\begin{gathered} x=\frac{-7-13}{4},\frac{-7+13}{4} \\ \\ x=\frac{-20}{4},\frac{6}{4} \\ \\ x=-5,\frac{3}{2} \end{gathered}[/tex]The solutions to the equation are: -5 and 3/2
SInce r is greater than s, we have:
r = 3/2
s = -5
Hence, to find r - s, let's subtract -5 from 3/2:
[tex]\begin{gathered} r-s=\frac{3}{2}-(-5) \\ \\ =\frac{3}{2}+\frac{5}{1} \\ \\ =\frac{3(1)+5(2)}{2} \\ \\ =\frac{3+10}{2} \\ \\ =\frac{13}{2} \end{gathered}[/tex]Therefore, the value of r - s is:
[tex]\frac{13}{2}[/tex]ANSWER: B
[tex]\frac{13}{2}[/tex]
