Question

If one zero of the polynomial 2x2 – 5x – (2k + 1) is twice the other, find both the zeros of the polynomial and the value of k.

Answer 1

Heya !!

Here's your answer..⬇⬇
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p(x) = 2x² - 5x - ( 2k + 1 )


$let \: \: \alpha \: \: be \: \: one \: \: zero. \\ \\ other \: \: zero \: \: will \: \: be \: \: 2 \alpha . \\ \\ sum \: \: of \: \: zeros \: = \frac{ - b}{a} \\ \\ \alpha + 2 \alpha = \frac{ - ( - 5)}{2} \\ \\ 3 \alpha = \frac{5}{2} \\ \\ \alpha = \frac{5}{6} \\ \\ product \: \: of \: \: the \: \: zeros \: = \frac{c}{a} \\ \\ \alpha \times 2 \alpha = \frac{ - (2k + 1)}{2} \\ \\ 2 {a}^{2} = \frac{ - (2k + 1)}{2} \\ \\ 2 {( \frac{5}{6} )}^{2} = \frac{ - (2k + 1)}{2} \\ \\ \frac{50}{36} = \frac{ - (2k + 1)}{2} \\ \\ 50 \times 2 = - 36(2k + 1) \\ \\ 100 = - 72k - 36 \\ \\ 100 + 36 = - 72k \\ \\ 136 = - 72k \\ \\ k = \frac{ - 136}{72} \\ \\ k = \frac{ - 17}{9} \\ \\ one \: \: zero \: = \alpha = \frac{5}{6} \\ \\ other \: \: zero \: \: = 2 \alpha = 2( \frac{5}{6} ) = \frac{10}{6}$
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Hope it helps..
Thanks :)

Answer 2

Answer:

$\text{The value of k is }\frac{-17}{9}\text{and its zeroes are }\frac{5}{6}\text{ and }\frac{5}{3}$

Step-by-step explanation:

$\text{Given that if one zero of the polynomial }2x^2- 5x -(2k + 1)\text{ is twice the other}$

$\text{we have to find the zeros of the polynomial and the value of k.}$

Let α is one zero of the polynomial.

As one zero is twice the other therefore other zero is 2α

$\text{sum of zeroes=}\alpha+2\alpha=\frac{-b}{a}=\frac{-(-5)}{2}=\frac{5}{2}$

$3\alpha=\frac{5}{2}$

$\alpha=\frac{5}{2\times 3}=\frac{5}{6}$

$Other zero=2\alpha=2\times \frac{5}{6}=\frac{5}{3}$

$\text{Product of zeroes=}\alpha.2\alpha=\frac{c}{a}=\frac{-(2k+1)}{2}$

$2(\alpha)^2=\frac{-(2k+1)}{2}$

$2\times 2\times (\frac{5}{6})^2=-2k-1$

$\frac{25}{9}+1=-2k$

$k=\frac{-17}{9}$

$\text{Hence, the value of k is }\frac{-17}{9}\text{and its zeroes are }\frac{5}{6}\text{ and }\frac{5}{3}$