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If sum of all zeros of the polynomial 5x2 – (3 + k)x + 7 is zero, then zeroes of the polynomial 2x2 – 2(k + 11)x + 30 are
  • a)
    3, 5
  • b)
    7, 9
  • c)
    3, 6
  • d)
    2, 5
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
If sum of all zeros of the polynomial 5x2 – (3 + k)x + 7 is zero...
Sum of zeroes of polynomial
5x2– (3 + k)x + 7 is 
According to question, ((3+k)/5) = 0 ⇒ k = –3
Now, 2x2 – 2(k + 11)x + 30 becomes 2x2 – 16x + 30.
i.e., 2x2 – 16x + 30 = 0 or x2 – 8x + 15 = 0
⇒ x = 3, 5
Hence, zeroes of polynomial 2x2 – 16x + 30 are 3, 5.
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Community Answer
If sum of all zeros of the polynomial 5x2 – (3 + k)x + 7 is zero...
If the sum of all zeros of the polynomial is 0, then the polynomial can be written as:

5x^2 = 0

To solve this equation, we can divide both sides by 5:

x^2 = 0

Taking the square root of both sides:

x = 0

Therefore, the only zero of the polynomial is 0.
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If sum of all zeros of the polynomial 5x2 – (3 + k)x + 7 is zero, then zeroes of the polynomial 2x2 – 2(k + 11)x + 30 area)3, 5b)7, 9c)3, 6d)2, 5Correct answer is option 'A'. Can you explain this answer?
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