Math, asked by s07849576, 1 year ago


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

Answers

Answered by MaheswariS
36

\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}

Answered by amitnrw
8

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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