Question

2x^2 + 7x - 15 = 0

If r and s are two solutions of the equation aboveand r>s, which of the following is thevalue of r-s?

Answer 1

Given the equation:

`2x^2+7x-15=0`

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:

`x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}`

Where:

a = 2

b = 7

c = -15

Thus, we have:

`\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=(-7\pm13)/(4) \\ \end{gathered}`

Solving further:

`\begin{gathered} x=(-7-13)/(4),(-7+13)/(4) \\ \\ x=(-20)/(4),(6)/(4) \\ \\ x=-5,(3)/(2) \end{gathered}`

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:

`\begin{gathered} r-s=(3)/(2)-(-5) \\ \\ =(3)/(2)+(5)/(1) \\ \\ =(3(1)+5(2))/(2) \\ \\ =(3+10)/(2) \\ \\ =(13)/(2) \end{gathered}`

Therefore, the value of r - s is:

`(13)/(2)`

ANSWER: B

`(13)/(2)`