Question

If the ratio of the roots of the equation ax2+bx+c=0 is r,then prove that (r+1)^2/r = b^2/ac

Answer 1

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Answer 2

Answer:

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

Step-by-step explanation:

If '$\alpha$' and '$\beta$'  are the roots of the equation  $ax^{2} +bx+c = 0$  

then we have,

Sum of roots = $\alpha + \beta = \frac{-b}{a}$  and Product of roots = $\alpha \beta = \frac{c}{a}$

Here given,  

ratio of the roots of the equation $ax^2+bx+c=0$ is 'r'

Then we have,

$\frac{\alpha }{\beta } = r$

Required to prove

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

LHS = $\frac{(r+1)^2}{r}$= $\frac{(\frac{\alpha }{\beta }+1)^2 }{\frac{\alpha }{\beta } }$

= $(\frac{\alpha +\beta }{\beta }) ^2 * \frac{\beta }{\alpha }$

=$\frac{(\alpha +\beta) ^2}{\alpha \beta }$

Substituting the value of $\alpha +\beta$ and $\alpha \beta$ in the above expression

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

= $\frac{b^2}{a^2} * \frac{a}{c}$

= $\frac{b^2}{ac}$ = RHS

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

Hence proved