if the zeros of the polynomial ax^2+bx+c be in ratio m:n then prove that b^2mn=(m^2+n^2)ac

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Solution

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$Let α and β are the roots of the quadratic equation {ax}^{2} + bx + c = 0 . Therefore, sum of roots (α + β) = \frac{b}{a} product of roots (αβ) = \frac{c}{a} It is given that the roots are in the ratio of m : n Therefore, α : β = m : n \Rightarrow \frac{α}{β} = \frac{m}{n} \Rightarrow \frac{α + β}{α - β} = \frac{m + n}{m - n} (by applying Componendo and Dividendo) \Rightarrow \frac{{(α + β)}^{2}}{{(α - β)}^{2}} = \frac{{(m + n)}^{2}}{{(m - n)}^{2}} (squaring both sides) \Rightarrow \frac{{(α + β)}^{2}}{{(α + β)}^{2} - 4 αβ} = \frac{{(m + n)}^{2}}{{(m - n)}^{2}} \Rightarrow \frac{{(\frac{b}{a})}^{2}}{{(\frac{b}{a})}^{2} - 4 (\frac{c}{a})} = \frac{{(m + n)}^{2}}{{(m - n)}^{2}} \Rightarrow \frac{b^{2}}{b^{2} - 4 ac} = \frac{{(m + n)}^{2}}{{(m - n)}^{2}} \Rightarrow b^{2} {(m - n)}^{2} = {(m + n)}^{2} (b^{2} - 4 ac) \Rightarrow b^{2} {(m - n)}^{2} = b^{2} {(m + n)}^{2} - 4 ac {(m + n)}^{2} \Rightarrow 4 ac {(m + n)}^{2} = = b^{2} {(m + n)}^{2} - b^{2} {(m - n)}^{2} \Rightarrow 4 ac {(m + n)}^{2} = = b^{2} ({(m + n)}^{2} - {(m - n)}^{2}) \Rightarrow 4 ac {(m + n)}^{2} = = b^{2} (4 mn) \Rightarrow ac {(m + n)}^{2} = = b^{2} (mn) \Rightarrow b^{2} (mn) = ac {(m + n)}^{2}$

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