Question

If the sum of zero and product of the zero of polynomial p (x)=x2 (k - 7)x (k 1) are equal than k

Answer 1

CORRECT QUESTION:-

If sum and products of the roots of the given polynomial p(x)=x²+(k-7)x+k+1 are equal then find the value of k

GIVEN:-

p(x)=x²+(k-7)x+k+1 and product and sum of the roots are equal

TO FIND:-

The value of k

EXPLANATION:-

$\underline{\underline{\bf SUM \: OF \: ROOTS:-}}$

We know that

$\boxed{\sf sum \: of \: roots =\dfrac{-coefficeint \: of \: x }{coefficient \: of \: x^2}=\dfrac{-b}{a}}$

here from the polynomial

b=(k-7)

a=1

c=k+1

Substituting

sum of roots =-(k-7)/1

$\underline{\underline{\bf PRODUCT \: OF \: ROOTS:-}}$

$\boxed{\sf product \: of \: roots =\dfrac{constant \: term }{coefficient \: of \: x^2}=\dfrac{c}{a}}$

Substituting

product of roots =k+1/1

Given

sum of roots=product of roots

=>-(k-7)=k+1

=>-k+7=k+1

=>2k=6

$\boxed{\sf k=3}$

Answer 2

Correct Question :

If the sum of zero and product of the zero of polynomial p(x) = x² + (k - 7)x + (k + 1) are equal. Then find the value of k.

AnswEr :

⋆ p(x) = x² + (k - 7)x + (k + 1)

• a = 1
• b = (k - 7)
• c = (k + 1)

• According to Question :

⇒ Sum of Zeros = Product of Zeros

⇒ $\sf{ \dfrac{Coefficient \: of \: x}{Coefficient of x^{2} } = \dfrac{Constant \: term}{Coefficient of x^{2} } }$

⇒ $\sf{ \dfrac{-b}{a} = \dfrac{c}{a} }$

⇒ $\sf{ \dfrac{-(k-7)}{1} = \dfrac{(k+1)}{1} }$

⇒ - (k - 7) = (k + 1)

⇒ - k + 7 = k + 1

⇒ 7 - 1 = k + k

⇒ 6 = 2k

⇒ $\mathsf{k = \cancel\dfrac{6}{2} }$

⇒ k = 3

⠀

჻ Therefore, Value of k is 3.