Given:
The equation is given as,
[tex]2x^2+7x-15=0[/tex]The solutions of the above equation are r and s.
The objective is to find r - s.
Explanation:
From the quadratic equation consider a = 2, b = 7 and c = -15.
The solutions r and s can be calculated using quadratic formula as,
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]Substitute the value of a, b and c in the above formula.
[tex]\begin{gathered} x=\frac{-7\pm\sqrt[]{7^2-4(2)(-15)}}{2(2)} \\ =\frac{-7\pm\sqrt[]{49+120}}{4} \\ =\frac{-7\pm\sqrt[]{169}}{4} \\ =\frac{-7\pm13}{4} \end{gathered}[/tex]To find r and s:
On further solving the equation,
[tex]\begin{gathered} x=\frac{-7+13}{4},\frac{-7-13}{4} \\ =\frac{6}{4},\frac{-20}{4} \\ =\frac{3}{2},-5 \end{gathered}[/tex]Since, it is given that r > s, the value of r is (3/2) and s is -5.
To find r - s:
Now, the difference can be calcualted as,
[tex]\begin{gathered} r-s=\frac{3}{2}-(-5) \\ =\frac{3}{2}+5 \\ =\frac{3}{2}+\frac{5(2)}{2} \\ =\frac{3}{2}+\frac{10}{2} \\ =\frac{13}{2} \end{gathered}[/tex]Thus, the value of r - s is (13/2).
Hence, option (B) is the correct answer.