Question

যদি ax²+ bx+c= 0, দ্বিঘাত সমীকরণের বীজদ্বয়ের অনুপাত 1: r হয়, তবে, দেখাও যে
(r+1)²/r=b²/ac​

Answer 1

$\textbf{Given:}$

$\text{Roots of the equation ax^2+bx+c=0 are in the ratio 1:r}$

$\textbf{To prove:}$

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

$\textbf{Solution:}$

$\text{Let the roots of ax^2+bx+c=0 be \alpha and \beta }$

$\text{As per given data,}$

$\alpha:\beta=1:r$

$\implies\,\alpha=k\;\text{and}\;\beta=kr$

$\text{Then,}$

$\alpha+\beta=\dfrac{-b}{a}$

$k+kr=\dfrac{-b}{a}$

$(1+r)k=\dfrac{-b}{a}$.......(1)

$\alpha\,\beta=\dfrac{c}{a}$

$k(kr)=\dfrac{c}{a}$

$k^2r=\dfrac{c}{a}$

$k^2=\dfrac{c}{ar}$.......(2)

$\text{Squaring equation (1), we get}$

$(1+r)^2k^2=\dfrac{b^2}{a^2}$

$\text{Using (2), we get}$

$(1+r)^2\dfrac{c}{ar}=\dfrac{b^2}{a^2}$

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

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

$\textbf{Hence proved}$

Answer 2

Given :

ax² + bx + c = 0,

the ratio of the two seeds of the quadratic equation is 1: r,  

To Find : show that (r + 1) ² / r = b² / ac

Solution :

Roots are α , αr

Sum of roots = α + αr   = - b/a

=> r + 1 = -b/aα

=> r  + 1 =    -b/aα  

Product of roots = α.αR   = c/a

=> r = c/aα²

(r + 1) ² / r = b² / ac

LHS =  (r + 1) ² / r

= ( -b/aα)² / (c/aα²)

= (b²/a²)/(c/a)

= b²/ac

= RHS

QED

(r + 1) ² / r = b² / ac

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