Question

)
is
7. If the sum of Zeroes and the pocduct
of zeros of the polynomial
p(x) = x² + (k-7)x + (k+1) are equal than k=
​

Answer 1

Given :

• p(x) = x² + (k-7)x + (k+1)

To find :

• the value of k

Formula used :

$\large{\sf \: s = \dfrac{ - b}{a} }$

$\bf p = \dfrac{c}{a}$

Where:-

• s = sum of roots
• p = product of roots
• b = coefficient of x
• a = coefficient of x²
• c = constant term

Solution :

x² + (k-7)x + (k+1) = 0

Here,

• a = 1
• b = k-7
• c = k+1

Now sum of roots :-

$\implies{\sf \: s = \dfrac{ - (k - 7)}{1} }$

$\implies{\sf \: s = - (k - 7)}$

$\implies{\sf \: s = - k + 7}$

$\implies{\sf \: s = 7 - k}$

Therefore sum of roots = 7-k

$\implies\sf p = \dfrac{c}{a}$

$\implies\sf p = \dfrac{k + 1}{1}$

$\implies\sf p = {k + 1}$

product of roots = k+1

It is given that product and sum of roots are equal. Therefore according to the question :-

$\implies{\sf \: 7 - k = k + 1}$

$\implies{\sf \: - k - k = - 7 + 1}$

$\implies{\sf \: - 2k = - 6}$

$\implies{\sf \: k = \dfrac{ - 6}{ - 2} }$

$\implies{\sf \: k = 3}$

Answer : 3