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If $r$ be the ratio of the roots of the equation $ax^{2}+bx+c=0,$ then 

  1. $\frac{r}{b}=\frac{r+1}{ac}$
  2. $\frac{r+1}{b}=\frac{r}{ac}$
  3. $\frac{(r+1)^{2}}{r}=\frac{b^{2}}{ac}$
  4. $\left(\frac{r}{b}\right)^{2}=\frac{r+1}{ac}$
in Numerical Ability by Boss (17.5k points)
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2 Answers

+2 votes

Answer: $\textbf C$

Let, $\alpha, \beta$ be the roots of the given equation, then we know that:

$\text{Sum of roots} \;\mathrm{=\alpha + \beta = \frac{-b}{a} }\tag{1}$

$\text{Product of roots}\;\mathrm{ = \alpha.\beta = \frac{c}{a}}\tag{2}$

$\mathbf{It \;is\; given\; that:}\;\frac{\alpha}{ \beta} = r \Rightarrow \alpha = r\beta \tag{3}$

Substituting $(3)$ in $(1)$, we get:

$\mathrm {r\beta + \beta = \frac{-b}{a}  \Rightarrow \beta(r+1) = \frac{-b}{a} \Rightarrow \beta=\frac{-b}{a(r+1)} }\tag{4}$
From $(2)$ and $(3)$, we get:

$\mathrm{=r\beta^2 = \frac{c}{a}}$

$\mathrm {\Rightarrow \frac{b^2}{a^2(r+1)^2}r=\frac{c}{a}}$

On simplifying, we get:

$\mathrm{\frac{(r+1)^2}{r} = \frac{b^2}{ac}}$

$\therefore \mathbf C$ is the correct answer.

by Boss (19.1k points)
edited by
+2
$r=\beta /\alpha$ is also possible.. but it won't matter :)
+2

@chirudeepnamini
Yeah. That's true.  It's because of the term $\frac{(r+1)^2}{r}$.

Let $f(r)=\frac{(r+1)^2}{r}$. It can be showed that $f(\frac{1}{r})=f(r)$.

Demonstration:

$f(\frac{1}{r})=\frac{(\frac{1}{r}+1)^2}{\frac{1}{r}}=\frac{\frac{(1+r)^2}{r^2}}{\frac{1}{r}}=\frac{(1+r)^2}{r^2}\times \frac{r}{1}=\frac{(r+1)^2}{r}=f(r)$

+1

@`JEET
This solution just needs basics of algebra. 👍 Upvoted.

+1
Thanks!
+1
@chirudeepnamini

Correct!
0 votes

We know the solution to the standard quadratic equation $ax^2+bx+c=0$ that
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. So the roots are $\frac{-b+\sqrt{b^2-4ac}}{2a}$ and $\frac{-b-\sqrt{b^2-4ac}}{2a}$.

 

According to the question,

$$\begin{align}\frac{\frac{-b+\sqrt{b^2-4ac}}{2a}}{\frac{-b-\sqrt{b^2-4ac}}{2a}}& =r \\ \Rightarrow \frac{-b+\sqrt{b^2-4ac}}{-b-\sqrt{b^2-4ac}} &=r \\ \Rightarrow \frac{2\sqrt{b^2-4ac}}{-2b} &=\frac{r-1}{r+1};~[\text{Dividendo-Compendo}] \\ \Rightarrow \frac{b^2-4ac}{b^2} &=\frac{(r-1)^2}{(r+1)^2};~[\text{Squaring both sides}]\\ \Rightarrow 1- \frac{4ac}{b^2}&= \frac{(r-1)^2}{(r+1)^2}\\ \Rightarrow 1-\frac{(r-1)^2}{(r+1)^2} &= \frac{4ac}{b^2}\\ \Rightarrow \frac{4r}{(r+1)^2}&=\frac{4ac}{b^2}; ~[\because (r+1)^2-(r-1)^2=r^2+2r+1-(r^2-2r+1)=4r]\\ \therefore \frac{(r+1)^2}{r}&=\frac{b^2}{ac}\end{align}$$

 

So the correct answer is C.

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