JEE Main 2026 — Quadratic Equation Question with Solution

Let $\alpha =a,\beta =ar,\gamma =a{r}^2,\delta =a{r}^3$ be the terms of the G.P.

From the given quadratic equations, the sum of the roots are:

$\alpha +\beta =1\Rightarrow a(1+r)=1$

$\gamma +\delta =4\Rightarrow a{r}^2(1+r)=4$

Dividing the second equation by the first equation:

$\frac{a{r}^2(1+r)}{a(1+r)}=\frac{4}{1}$

$r^2=4\Rightarrow r=2$ or $r=-2$

If $r=2$, then $a(1+2)=1\Rightarrow a=\frac{1}{3}$.

The product of the roots of the first equation is $p=\alpha \beta =a^{2}r={\left(\frac{1}{3}\right)}^2\times 2=\frac{2}{9}$. Since $p\in \mathbf{Z}$, $r=2$ is rejected.

If $r=-2$, then $a(1-2)=1\Rightarrow a=-1$.

The product of the roots of the first equation is $p=\alpha \beta =a^{2}r=(-1{)}^2\times (-2)=-2$.

The product of the roots of the second equation is $q=\gamma \delta =(a{r}^2)(a{r}^3)=a^2{r}^5=(-1{)}^2\times (-2{)}^5=-32$.

Both $p$ and $q$ are integers, which satisfies the given condition.

Therefore, $|p+q|=|-2-32|=|-34|=34$.

Answer: $34$