Question

if one zero of the polynomial 2x^2-5x-(2k+1) is twice the other,find both the zeroes of polynomial and 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:

Hii User!!

Let, p(x)=2x²-5x-(2k+1). =) eq (i)

Also let, α be one root of equation (i)

=)Then the other root will be 2α.

=)α+2α=-(-5)/2

=) 3α=5/2

=) α=5/6

Here ,α×2α=-(2k+1)/2

=) 2α²=-(2k+1)/2

=) -(2k+1)/2=2×(5/6)²

=) -(2k+1)=4×25/36

=)-2k-1=25/9

= -2k=25/9+1

=) -2k=34/9

=)  k=-34/9×1/2

=) k=-17/9

Hence, the value of k is -17/9