An algebra problem by A Former Brilliant Member

Using Vieta's formula: $x^2-px+r=0\small{\begin {cases} \color{#D61F06}{\alpha+\beta=p} \\ \alpha\beta=r\end {cases}}$ $x^2-qx+r=0\small{ \begin {cases}\dfrac{\alpha}2+2\beta=q\implies \color{#D61F06}{\alpha+4\beta=2q }\\ \alpha\beta=r\end {cases}}$

Solving $\color{#D61F06}{\alpha+\beta=p}$ and $\color{#D61F06}{\alpha+4\beta=2q }$ for $\alpha$ and $\beta$ , we get:

$\alpha= \dfrac{2(2p-q)}3, \beta=\dfrac{2q-p}3$

$\therefore r=\alpha\beta=\dfrac{2(2q-p)(2p-q)}9$

$\LARGE\therefore \dfrac 34\times \dfrac{2}9=\dfrac 16$

$\huge \therefore 1+6=\boxed 7$

$x^2-px+r=0$ has roots $\alpha$ and $\beta$ . From Vieta's formula, we have

$\alpha+\beta=p\\ \alpha\beta=r$

$x^2-qx+r=0$ has roots $\dfrac{\alpha}{2}$ and $2\beta$ . From Vieta's formula:

$\dfrac{\alpha}{2}+2\beta=q\\ \alpha\beta=r$

Given that

$\dfrac{3r}{4}=k(2p-q)(2q-p)\\ \dfrac{3\alpha\beta}{4}=k\Big(2(\alpha+\beta)-(\dfrac{\alpha}{2}+2\beta)\Big)\Big(2(\dfrac{\alpha}{2}+2\beta)-(\alpha+\beta)\Big)\\ \dfrac{3\alpha\beta}{4}=k\Big(2\alpha+2\beta-\dfrac{\alpha}{2}-2\beta\Big)\Big(\alpha+4\beta-\alpha-\beta\Big)\\ \dfrac{3\alpha\beta}{4}=k\Big(\dfrac{3\alpha}{2}\Big)\Big(3\beta\Big)\\ k=\left(\dfrac{\color{#3D99F6}{\cancel{\color{#333333}{3\alpha}}}\color{#EC7300}{\cancel{\color{#333333}{\beta}}}}{\color{#D61F06}{\cancel{\color{#333333}{4}}2}}\right)\left(\dfrac{\color{#D61F06}{\cancel{\color{#333333}{2}}}}{\color{#3D99F6}{\cancel{\color{#333333}{3\alpha}}}}\right)\left(\dfrac{1}{3\color{#EC7300}{\cancel{\color{#333333}{\beta}}}}\right)\\=\dfrac{1}{6}$

$a=1,\;b=6,\;a+b=1+6=\boxed{7}$

Using Vieta formulas , we have:

$\begin{cases} \alpha + \beta = p & ...(1) \\ \dfrac \alpha 2 + 2 \beta = q & ...(2) \\ \alpha \beta = r & ...(3) \end{cases}$

$\begin{aligned} 2(2)-(1): \quad 3 \beta & = 2q = q \\ \implies \beta & = \frac 13 (2q-p) \end{aligned}$

$\begin{aligned} (1): \quad \alpha & = p - \frac 13 (2q-p) \\ & = \frac 23 (2p-q) \end{aligned}$

$\begin{aligned} \frac 34 (3): \quad \frac 34 r & = \frac 34 \alpha \beta \\ & = \frac 34 \cdot \frac 23 (2q-p) \cdot \frac 13 (2q-p) \\ & = \frac 16 (2p-q)(2q-p) \end{aligned}$

$\implies a+b = 1 +6 = \boxed{7}$