PSAT #1, Section 3, #6

We’re told that r and s are the two solutions to 2x^2+7x-15=0, so in order to solve for r-s, we’ll need to solve that quadratic. Doesn’t look like the EASIEST quadratic to factor, so I’m gonna jump straight to the quadratic formula.

    \begin{align*}x&=\dfrac{-7\pm\sqrt{7^2-4(2)(-15)}}{2(2)}\\x&=\dfrac{-7\pm\sqrt{169}}{4}\\x&=-\dfrac{7}{4}\pm\dfrac{13}{4}\end{align*}

Because the question tells you that r>s, you can say that r=-\dfrac{7}{4}+\dfrac{13}{4}=\dfrac{6}{4}=\dfrac{3}{2} and s=-\dfrac{7}{4}-\dfrac{13}{4}=-\dfrac{20}{4}=-5. Therefore,

    \begin{align*}r-s&=\dfrac{3}{2}-(-5)\\&=\dfrac{3}{2}+\dfrac{10}{2}\\&=\dfrac{13}{2}\end{align*}

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