Question

The value of k such that 3x2+2kx-k-5 has the sum of the zeroes as half of their product is​

Answer 1

$\huge \bf\green{ \boxed{\bf \orange{Answer}}} \pink \downarrow$

$\bf{Let \:  P \: (x) \: = \: {3x}^{2} \: + \: 2kx \: + \: x \: − \: k \: − \: 5}$

$\bf{= \: {3x}^{2} \: + \: x(2k+1) \: − \: (k \: + \: 5)}$

$\bf{i.e;  {ax}^{2} \: +bx \: + \: c}$

$\bf{a \: = \: 3; \: b \: = \: 2k \: + \: 1 \: ; \: c \: = \: − \: (k \: + \: 5)}$

$\bf{1) \: Sum \: of \: zeroes \: = \: \frac{ - b}{a} \: = \: \frac{ - (2k + 1)}{3}}$

$\bf{2) \: Product \: of \: zeroes \: = \: \frac{c}{a} \: = \: \frac{ - (k + 5)}{3}}$

$\huge\bf{According \: to \: given }$

$\bf{−(2k+1)/3= \frac{1}{2} [−(k+5)/3]}$

$\bf{2(2k+1)=k+5}$

$\bf{4k+2=k+5}$

$\bf{3k=3⇒k=1}$

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