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Let $\alpha$ and $\beta$ be the roots of the quadratic equation $a x^2+b x+c=0$. Observe the lists given below
The correct match of List-I from List-II is
(i) (ii) (iii) (iv)
Options:
The correct match of List-I from List-II is
(i) (ii) (iii) (iv)
Solution:
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Verified Answer
The correct answer is:
E $\quad$ B $\quad$ D $\quad$ A
Using the condition that the roots of $a x^2+b x+c=0$ may be in the ratio $m: n$ is $m n b^2=a c(m+n)^2$.
(i) If the roots are $\alpha=\beta$, then
$$
\begin{aligned}
\alpha \cdot \alpha b^2 & =a c(\alpha+\alpha)^2 \\
\Rightarrow \quad b^2 & =4 a c
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \quad b^2=4 a c \\
& \text { (ii) If the roots are } \alpha=2 \beta \text {, then } \\
& \\
& \Rightarrow \quad \beta \cdot 2 \beta b^2=a c(\beta+2 \beta)^2 \\
& 2 b^2=9 a c
\end{aligned}
$$
(iii) If the roots are $\alpha=3 \beta$, then
$$
\Rightarrow \quad \begin{aligned}
\beta \cdot 3 \beta b^2 & =a c(\beta+3 \beta)^2 \\
\Rightarrow \quad 3 b^2 & =16 a c
\end{aligned}
$$
(iv) If the roots are $\alpha=\beta^2$, then
$$
\begin{aligned}
& \left(a^2 c\right)^{\frac{1}{2+1}}+\left(a c^2\right)^{\frac{1}{2+1}}=-b \\
\Rightarrow \quad & \left(a^2 c\right)^{\frac{1}{3}}+\left(a c^2\right)^{\frac{1}{3}}=-b
\end{aligned}
$$
Hence, option (d) is correct.
(i) If the roots are $\alpha=\beta$, then
$$
\begin{aligned}
\alpha \cdot \alpha b^2 & =a c(\alpha+\alpha)^2 \\
\Rightarrow \quad b^2 & =4 a c
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \quad b^2=4 a c \\
& \text { (ii) If the roots are } \alpha=2 \beta \text {, then } \\
& \\
& \Rightarrow \quad \beta \cdot 2 \beta b^2=a c(\beta+2 \beta)^2 \\
& 2 b^2=9 a c
\end{aligned}
$$
(iii) If the roots are $\alpha=3 \beta$, then
$$
\Rightarrow \quad \begin{aligned}
\beta \cdot 3 \beta b^2 & =a c(\beta+3 \beta)^2 \\
\Rightarrow \quad 3 b^2 & =16 a c
\end{aligned}
$$
(iv) If the roots are $\alpha=\beta^2$, then
$$
\begin{aligned}
& \left(a^2 c\right)^{\frac{1}{2+1}}+\left(a c^2\right)^{\frac{1}{2+1}}=-b \\
\Rightarrow \quad & \left(a^2 c\right)^{\frac{1}{3}}+\left(a c^2\right)^{\frac{1}{3}}=-b
\end{aligned}
$$
Hence, option (d) is correct.
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