Question

Q2.If the sum of the zeroes of the polynomial p(x) = kx' +2x+3k is equal to their product, then
find two value of 'k'.​ .​

Answer 1

Answer:

Explanation:

Given :

• Polynomial, p(x) = kx²+2x+3k.
• Sum of zeroes = Product of zeroes.

To Find :

• The value of k.

Solution :

Given, polynomial, p(x) = kx²+2x+3k.

On comparing with, ax² + bx + c, We get ;

⇒ a = k , b = 2 , c = 3k

★ Sum of zeroes,

⇒ α + β = -b/a

⇒ α + β = -2/k

★ Product of zeroes,

⇒ αβ = c/a

⇒ αβ = 3k/k

⇒ αβ = 3

Given, sum of zeroes = product of zeroes.

⇒ -2/k = 3

⇒ -2 = 3k

⇒ k = -2/3

Hence, The value of k is -2/3.

Answer 2

⠀⠀⠀⠀⠀⠀${\huge{\underbrace{\rm{Question}}}}$

Q2.If the sum of the zeroes of the polynomial p(x) = kx² +2x+3k is equal to their product, then find two value of 'k'.

⠀⠀⠀⠀⠀⠀⠀${\huge{\underbrace{\rm{Answer}}}}$

⠀

Given:

⠀

• sum of the zeroes of the polynomial p(x) = kx' +2x+3k is equal to their product

⠀

To find:

⠀

find the value of 'k'.

⠀

Solution:

⠀

We have, p(x) = kx² +2x+3k

Comparing the given polynomial p(x) = kx² +2x+3k with the standard equation ax² + bx + c , we get,

⠀

• a = k

• b = 2

• c = 3k

⠀

We know that,

⠀

⠀⠀$\sf{Sum\:of\:zeroes=\alpha\:+\beta=\dfrac{-b}{a}}$

⠀

⠀⠀⠀⠀⠀⠀$\pink\bigstar$ $\sf{\alpha+\beta=\dfrac{-2}{k}}$

⠀

⠀⠀⠀$\sf{Product\:of\:zeroes=\alpha×\beta=\dfrac{c}{a}}$

⠀

⠀⠀⠀⠀⠀⠀$\pink\bigstar$ $\sf{\alpha\beta=\dfrac{3k}{k}=3}$

⠀

⠀⠀⠀

⠀⠀⠀⠀⠀$\begin{gathered}\begin{gathered}\bf{\dag}\;{\underline{\sf{\pink{According\:to\:the\:question}}}}\\ \\\end{gathered} \end{gathered}$

⠀

⠀⠀⠀⠀⠀⠀⠀$\sf{:\implies \alpha+\beta=\alpha\beta}$

⠀

⠀⠀⠀⠀⠀⠀⠀$\sf{:\implies \dfrac{-2}{k}=3}$

⠀⠀⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀$\sf{:\implies 3k=-2}$

⠀⠀

⠀⠀⠀⠀⠀⠀⠀$\begin{gathered}\begin{gathered}:\implies{\underline{\boxed{\sf{\purple{k=\dfrac{-2}{3}}}}}}\;\bigstar\\ \\\end{gathered}\end{gathered}$

⠀⠀⠀⠀⠀$\therefore\:{\underline{\sf{The\:value\:of\:k\:is\:\dfrac{-2}{3}}}}$

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