if alpha and beta are the zeros of the polynomial find this answer please
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2x² - 5x + k = 0
Sum of zeroes = ã ( alpha) + ß(beta) = - b / a
= 5 / 2
It is given that ã - ß = 5
So ã + ß = 5/2
2(ã + ß) = 5
2ã + 2ß = 5 ...(1)
and ã - ß = 5 ..(2)
Therefore ã = 5 + ß ...(3)
put the value of ã in equation in equation (1).
2 ( 5 + ß ) + 2ß = 5
10 + 2ß + 2ß = 5
4ß = 5 - 10
4ß = -5
ß = -5/4
put the value of ß in equation (3).
ã = 5 + (-5/4)
= 5 - 5/4
= 20 - 5 /4
= 15 /4
Now product of zeroes = ãß = c/a
k /2 = 15/4 × -5 /4
k / 2 = ( - 75 / 16)
k = (- 75 / 16) × 2
k = (- 75 / 8)
Hope you like it please mark as brainliest and follow me if you like my answer.
Sum of zeroes = ã ( alpha) + ß(beta) = - b / a
= 5 / 2
It is given that ã - ß = 5
So ã + ß = 5/2
2(ã + ß) = 5
2ã + 2ß = 5 ...(1)
and ã - ß = 5 ..(2)
Therefore ã = 5 + ß ...(3)
put the value of ã in equation in equation (1).
2 ( 5 + ß ) + 2ß = 5
10 + 2ß + 2ß = 5
4ß = 5 - 10
4ß = -5
ß = -5/4
put the value of ß in equation (3).
ã = 5 + (-5/4)
= 5 - 5/4
= 20 - 5 /4
= 15 /4
Now product of zeroes = ãß = c/a
k /2 = 15/4 × -5 /4
k / 2 = ( - 75 / 16)
k = (- 75 / 16) × 2
k = (- 75 / 8)
Hope you like it please mark as brainliest and follow me if you like my answer.
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