Question

please answer me this 6th question class 10 ​

Answer 1

Question:

Find the value of k in the given polynomial such that 3 becomes zero of the polynomial p(x) = 2x^2 - 3kx + 2

Answer:

• k = 20/9

Explanation:

Let's say 3 is a zero of p(x)

⇒ x = 3

∴ p(3) = 2(3)² - 3k(3) + 2 = 0

Solving for k :-

⇒ 2(3)² - 3k(3) + 2 = 0

⇒ 2(9) - 9k + 2 = 0

⇒ 18 - 9k + 2 = 0

⇒ 20 - 9k = 0

⇒ 20 = 9k

⇒ 20/9 = k

∴ The value of ‘k’ will be 20/9

Hence, we can say that the value of k would be 20/9 such that 3 would be a zero of p(x) = 2x² - 3kx + 2.

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Answer 2

Question :-

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Find the value of k in the given polynomial such that 3 becomes zero of the polynomial p(x) = 2x² - 3kx + 2.

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Solution :-

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Let 3 be the zero of polynomial p(x) where, x is equal to 3.

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Therefore, according to the question, we get :-

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$\\ \longrightarrow \sf 2 \times {3}^{2} - 3k(3) + 2 = 0$

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Now, we will solve for the value of "k", we get :-

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$\\ \sf \longrightarrow 2 \times 9 - 9k + 2 = 0 \\ \\ \\ \sf \longrightarrow 18 + 2 = 9k \\ \\ \\ \sf \longrightarrow 20 = 9k \\ \\ \\ \star \: \purple{\underline{ \boxed{ \frak{ k = \frac{20}{9} }}}} \: \star$

Therefore, the value of k is 20/9.

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