Please help answer question 15
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Given Polynomial is (a^2 + 9)x^2 + 13x + 6a.
Let one zero of the Quadratic polynomial be a.
Given one of zero of the Quadratic polynomial is reciprocal of the other = 1/a.
Now,
We know that sum of zeroes = -b/a
= > a + (1/a) = -13/a
= > a + (1/a) = -13/(a^2 + 9)
We know that product of zeroes = c/a
= > a * (1/a) = 6a/(a^2 + 9)
= > 1 = 6a/a^2 + 9
= > a^2 + 9 = 6a
= > a^2 - 6a + 9 = 0
= > (a - 3)^2 = 0
= > a - 3 = 0
= > a = 3.
Therefore the value of a = 3.
Hope this helps!
Let one zero of the Quadratic polynomial be a.
Given one of zero of the Quadratic polynomial is reciprocal of the other = 1/a.
Now,
We know that sum of zeroes = -b/a
= > a + (1/a) = -13/a
= > a + (1/a) = -13/(a^2 + 9)
We know that product of zeroes = c/a
= > a * (1/a) = 6a/(a^2 + 9)
= > 1 = 6a/a^2 + 9
= > a^2 + 9 = 6a
= > a^2 - 6a + 9 = 0
= > (a - 3)^2 = 0
= > a - 3 = 0
= > a = 3.
Therefore the value of a = 3.
Hope this helps!
siddhartharao77:
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Answered by
1
Given Polynomial is (a^2 + 9)x^2 + 13x + 6a.
Let one zero of the Quadratic polynomial be a.
Given one of zero of the Quadratic polynomial is reciprocal of the other = 1/a.
Now,
We know that sum of zeroes = -b/a
= > a + (1/a) = -13/a
= > a + (1/a) = -13/(a^2 + 9)
We know that product of zeroes = c/a
= > a * (1/a) = 6a/(a^2 + 9)
= > 1 = 6a/a^2 + 9
= > a^2 + 9 = 6a
= > a^2 - 6a + 9 = 0
= > (a - 3)^2 = 0
= > a - 3 = 0
= > a = 3.
Hope this helps!
Let one zero of the Quadratic polynomial be a.
Given one of zero of the Quadratic polynomial is reciprocal of the other = 1/a.
Now,
We know that sum of zeroes = -b/a
= > a + (1/a) = -13/a
= > a + (1/a) = -13/(a^2 + 9)
We know that product of zeroes = c/a
= > a * (1/a) = 6a/(a^2 + 9)
= > 1 = 6a/a^2 + 9
= > a^2 + 9 = 6a
= > a^2 - 6a + 9 = 0
= > (a - 3)^2 = 0
= > a - 3 = 0
= > a = 3.
Hope this helps!
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