CBSE Class 10 Answered
if underroot 2 and -underroot 2
are the zeroes of p(x) = 2x^$ + 7x^3 - 19x^2 - 14x + 30, then find the other zeroes of p(x)
Asked by | 21 Jul, 2012, 06:55: PM
Expert Answer
Answer : Given : If underroot 2 and -underroot 2 are the zeroes of p(x) = 2x4 + 7x3 - 19x2 - 14x + 30
To find : the other zeroes of p(x)
Since we know that -underrroot 2 is a zero of p(x) , p(-underrroot 2) = 0
=> x- (-underrroot 2 ) is a factor of p(x)
dividing 2x4 + 7x3 - 19x2 - 14x + 30 by x-(-underrroot 2) i.e x+underrroot 2, we get
we get the quotient = 2x3 + (7 - (2 * ( underroot 2 ) ) ) x2 - (15 + (7*underroot 2))x + 15 * underrroot 2
=> 2x4 + 7x3 - 19x2 - 14x + 30 = (x+underrroot 2)
* (2x3 + (7 - (2 * ( underroot 2 ) ) ) x2 - (15 + (7*underroot 2))x + 15 * underrroot 2 )
Also underroot 2 is a zero of of 2x3 + (7 - (2 * ( underroot 2 ) ) ) x2 - (15 + (7*underroot 2))x + 15 * underrroot 2
therefore underroot 2 is the factor of 2x3 + (7 - (2 * ( underroot 2 ) ) ) x2 - (15 + (7*underroot 2))x + 15 * underrroot 2 =0
By dividing 2x3 + (7 - (2 * ( underroot 2 ) ) ) x2 - (15 + (7*underroot 2))x + 15 * underrroot 2 by (x - underroot 2 ) , the quotient is 2x2 + 7x - 15
Therefore p(x) i.e 2x4 + 7x3 - 19x2 - 14x + 30 = (x + underoot 2 ) * (x -underoot 2 ) * ( 2x2 + 7x - 15 )
and 2x2 + 7x - 15 = ( 2x - 3 ) ( x + 5)
=> (2x4 + 7x3 - 19x2 - 14x + 30 ) = (x + underoot 2 ) * (x -underoot 2 ) * ( 2x - 3 ) * ( x+5 )
=> All the zeroes of p(x) are
underoot 2 , -underoot 2 , 3/2 , -5
=> The other roots are 3/2 and -5 Answer
Answered by | 22 Jul, 2012, 12:40: AM
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