IMO Practice Test — Quadratic Equations

Explanation: Let the roots be $2k$ and $3k$. Sum $=5k=-p \Rightarrow k=-\frac{p}{5}$. Product $=6k^{2}=q \Rightarrow 6 \cdot \frac{p^{2}}{25}=q \Rightarrow 6p^{2}=25q$.

Explanation: Let $\sqrt{x}=p$, so $\sqrt{x+5}=5-p$. Squaring: $x+5=25-10p+p^{2}$. Since $p^{2}=x$, this gives $5=25-10p \Rightarrow p=2 \Rightarrow x=4$. Check: $\sqrt{4}+\sqrt{9}=2+3=5$.

Explanation: Let the roots be $\alpha$ and $2\alpha$. Sum $=3\alpha=6 \Rightarrow \alpha=2$, so the roots are $2$ and $4$. Then $k=$ product $=2 \times 4 = 8$.

Explanation: If the roots are reciprocals, their product is $1$. For $3x^{2}-10x+k=0$ the product of roots is $\frac{k}{3}$, so $\frac{k}{3}=1 \Rightarrow k=3$ (the roots are $3$ and $\frac{1}{3}$).

Explanation: Let the common root be $\alpha$. Subtracting the two equations gives $(p-p')\alpha+(q-q')=0$, so $\alpha=\frac{q'-q}{p-p'}$. Substituting back and simplifying gives the condition $(q-q')^{2}=(p-p')(p'q-pq')$.

Explanation: Let y=\(x^{2}\): \(2y^{2}-5y+3\)=0→\(2y^{2}-2y-3y+3\)=0→2y(y−1)−3(y−1)=0→(y−1)(2y−3)=0→y=1 or y=3/2. So \(x^{2}\)=1→x=±1; \(x^{2}\)=3/2→x=±\(\sqrt{3/2}\)