Question

Find the value of K such that the polynomial
x²- (K+6)x + 2(2k-1) has sum of its
zeroes equal to half of their product​

Answer 1

Given-

• $\bf{x}^{2} - (k + 6)x + 2(2k - 1)$
• product of this equation=2sum of this equation

To Find-

• value of k

Concept used-

• $\bf \: \alpha + \beta = \frac{ - b}{a}$
• $\bf \: \alpha \beta = \frac{c}{a}$

Solution-

We know

$\bf \: \alpha + \beta = \frac{ - b}{a} \\ \\ \bf \: here \: b = - (k + 6) \: \: a = 1 \\ \\ \bf \implies \alpha + \beta = \frac{k + 6}{1} \\ \\ \boxed{ \bf \implies \alpha + \beta = k + 6}$

and also

$\bf \: \alpha \beta = \frac{c}{a} \\ \\ \bf \: here \: c = 2(2k - 1) \: and \: a = 1 \\ \\ \implies \bf \alpha \beta = \frac{2(2k - 1)}{1} \\ \\ \implies \boxed{ \bf \: \alpha \beta = 2(2k - 1)}$

We have given that

$\bf \: \alpha \beta = 2 (\alpha + \beta ) \\ \\ \implies \bf 2(2k -1 ) = 2(k + 6) \\ \\ \implies \bf \: 2k - 1 = k + 6 \: \: \: \: \: \: \: \: \: \: \\ \\ \implies \bf \: 2k - k = 6 + 1 \: \: \: \: \: \: \: \: \: \\ \\ \implies \underline{ \boxed{ \huge \bf k = 7}}$

So the value of k is 7