Q4 and 5
Friends Pls help...
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help me also with this answer
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Hello sister ✌
Here's your answer friend,
⏩ Given : Two zeroes of P(x) = 3x4 - 12x³ + 5x² +16x -12 = 0 are -2/√3 or 2/√3.
Therefore,
[ X + 2/√3][X - 2/√3] = 0
[ (X)² - (2/√3)² ]
==> X² - 4/3
therefore,
on dividing P(x) by X² - 4/3 we get,
X² - 4/3 = 0
==> it can be written as 3x² - 4
Division is attached with a picture... ✌
on dividing p(x) by 3x² -4 we get,
x² -4x + 3 = 0
==> x² -3x -x + 3 = 0
==> x(x -3) -1( x - 3) = 0
==> (x - 1 )(x - 3) =0
==> x = 1 or x = 3
i. e. Zeroes of above the polynomial are 1, 3, 2/√3 and -2/√3.
Here's your answer friend,
⏩ Given : Two zeroes of P(x) = 3x4 - 12x³ + 5x² +16x -12 = 0 are -2/√3 or 2/√3.
Therefore,
[ X + 2/√3][X - 2/√3] = 0
[ (X)² - (2/√3)² ]
==> X² - 4/3
therefore,
on dividing P(x) by X² - 4/3 we get,
X² - 4/3 = 0
==> it can be written as 3x² - 4
Division is attached with a picture... ✌
on dividing p(x) by 3x² -4 we get,
x² -4x + 3 = 0
==> x² -3x -x + 3 = 0
==> x(x -3) -1( x - 3) = 0
==> (x - 1 )(x - 3) =0
==> x = 1 or x = 3
i. e. Zeroes of above the polynomial are 1, 3, 2/√3 and -2/√3.
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Damini1111:
Thanks a lot...... di for ur answer
It's my pleasure to help you sister : )
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