Question

One zero of polynomial 2x2 - 5x - (2k + 1) is double than second zero. Find out both zeroes and K.
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Answer 1

$\tt\blue{\underline{Answer:}}$

$\green{\tt\therefore \alpha= \frac{5}{3} }$

$\green{\tt \therefore\beta = \frac{5}{6} }$

$\green{\tt \therefore k = \frac{-17}{9} }$

$\tt \green { \underline{ Given : }} \\ \tt: \implies Eqn = 2x^{2} - 5x - (2k + 1) \\ \\ \tt \red { \underline{ To \: Find : }} \\ \tt: \implies Value \: of \: zeroes = ? \\ \\ \tt: \implies Value \: of \: k = ?$

$\tt\orange{\underline{Step-by-step\:\:explanation:}}$

• According to given question :

$\sf Let \: two \: zeroes \: be \: \alpha \: and \: \beta \\ \\ \tt\because \alpha = 2 \beta \: \: \: (Given) \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies \alpha + \beta = \frac{ - b}{a} \\ \\ \tt: \implies \alpha + \beta = \frac{ - ( - 5)}{2} \\ \\ \tt: \implies \alpha + \beta = \frac{5}{2} \\ \\ \tt: \implies 2 \beta + \beta = \frac{5}{2} \\ \\ \tt: \implies 3\beta = \frac{5}{2} \\ \\ \green{\tt: \implies \beta = \frac{5}{6}} \\ \\ \sf{putting \: value \: of \: \beta } \\ \tt: \implies \alpha + \frac{5}{6} = \frac{5}{2} \\ \\ \tt: \implies \alpha = \frac{5}{2} - \frac{5}{6} \\ \\ \tt: \implies \alpha = \frac{15 - 5}{6} \\ \\ \green{\tt: \implies \alpha = \frac{10}{6}=\frac{5}{3}} \\ \\ \bold{Again :} \\ \tt: \implies \alpha \beta = \frac{c}{a} \\ \\ \tt: \implies \frac{10}{6} \times \frac{5}{6} = \frac{ - (2k + 1)}{2} \\ \\ \tt: \implies \frac{50}{36} = \frac{ - 2k - 1}{2} \\ \\ \tt: \implies 50 = 18( - 2k - 1) \\ \\ \tt: \implies 50 = - 36k - 18 \\ \\ \tt: \implies 36k =- 50 - 18 \\ \\ \tt: \implies 36k = -68 \\ \\ \green{\tt: \implies k = \frac{-68}{36} = \frac{-17}{9} }$

Answer 2

$\Huge\textrm{\underline{Answer}}$

$\rm\large\rightarrow k=-\dfrac{17}{9}$

$\rm\large\rightarrow x=\dfrac{5}{6}\textrm{ or }x=\dfrac{5}{3}$

$\Huge\textrm{\underline{Solution}}$

$\textrm{Given polynomial,}$

$\rm\rightarrow2x^2-5x-(2k+1)$

$\textrm{Let two zeros be }\rm{x=\alpha,\beta.}$

$\rm\rightarrow \beta=2\alpha$

$\rm\text{The sum of the zeros is }-\dfrac{b}{a}.$

$\rm{\rightarrow}3\alpha=-\dfrac{-5}{2}$

$\rm{\rightarrow}\alpha=\dfrac{5}{6}$

$\textrm{Hence,}$

$\rm\rightarrow\red{\underline{\begin{cases} & \alpha=\dfrac{5}{6} \\ & \beta=\dfrac{5}{3} \end{cases}}\tiny{\text{//}}}$

$\rm\text{The product of the zeros is }\dfrac{c}{a}.$

$\rm\rightarrow\dfrac{5}{6}\times\dfrac{5}{3}=\dfrac{-(2k+1)}{2}$

$\rm\rightarrow\dfrac{25}{9}=-(2k+1)$

$\rm\rightarrow9(2k+1)=-25$

$\rm\rightarrow18k+9=-25$

$\rm\rightarrow18k=-34$

$\rm\rightarrow \red{\underline{k=-\dfrac{17}{9}}\tiny{\text{//}}}$

$\Huge\textrm{\underline{Verification}}$

$\textrm{Required polynomial,}$

$\rm\rightarrow2x^2-5x-(2k+1)$

$\rm=2x^2-5x-\left(-\dfrac{34}{9}+1\right)$

$\rm=2x^2-5x-\left(-\dfrac{25}{9}\right)$

$\rm=2x^2-5x+\dfrac{25}{9}$

$\textrm{The sum of zeros}$

$\rm\rightarrow-\dfrac{-5}{2}=\dfrac{5}{2}$

$\rightarrow\dfrac{5}{6}+\dfrac{5}{3}=\dfrac{5}{2}$

$\textrm{The product of zeros}$

$\rightarrow\dfrac{\frac{25}{9}}{2}=\dfrac{25}{18}$

$\rightarrow\dfrac{5}{6}\times\dfrac{5}{3}=\dfrac{25}{18}$

$\large\textrm{{\underline{Hence verified.}}}$