Vieta's formula made easy

Use the quadratic formula

$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{46\pm\sqrt{(-46)^2-4(15)(34)}}{2(15)}=\dfrac{23}{15}\pm\dfrac{\sqrt{19}}{15}$

It follows that $x_1=\dfrac{23+\sqrt{19}}{15}$ and $x_2=\dfrac{23-\sqrt{19}}{15}$

Adding their reciprocals, we obtain

$\dfrac{15}{23+\sqrt{19}}+\dfrac{15}{23-\sqrt{19}}$

$=\dfrac{15(23-\sqrt{19})+15(23+\sqrt{19})}{(23+\sqrt{19})(23-\sqrt{19})}$

$=\dfrac{345-15\sqrt{19}+345+15\sqrt{19}}{529-23\sqrt{19}+23\sqrt{19}-19}$

$=\dfrac{690}{510}$

$=\dfrac{23}{17}$

Finally,

$p+q=23+17=$ $\color{#3D99F6}\large\boxed{40}$ $\boxed{\text{answer}}$

Vieta's formula states that for a quadratic $ax^2+bx+c$ , the sum of the roots can be described as $\frac{-b}{a}$ and the product of the roots can be described by $\frac{c}{a}$ .

Let the roots of a quadratic be $p$ and $q$ . Therefore, the sum of the reciprocals of the roots must be $\frac{1}{p}$ + $\frac{1}{q}$ . Simplifying yields $\frac{p+q}{pq}$ .

In this case, the quadratic is $15x^2-46x+34$ , so the sum of the reciprocals of the roots must be $\frac{-b/a}{c/a}$ or $\frac{-b}{c}$ , or $\frac{46}{34}$ . Simplifying yields $\frac{23}{17}$ , and $23+17$ = $\boxed{40}$