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If the squared difference of the zeroes of the quadratic equation polynomial F(x)=x2+px+45 is equal to 144,find the value of 'p'.

Posted by Shriyanshi Gupta 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Abhinav Dutta 6 years, 10 months ago

Let the zerors be x and y Sum of zeros= x+y = -p Product of zeroes= xy = 45 A.T.Q ( x-y)2 = 144 x2 + y2 -2xy = 144 We write x2+y2 as (x+y)2 -2xy (x+y)2 -2xy -2xy = 144 (x+y)2 -4xy = 144 On putting above values (-p)2 -4(45) = 144 p2 -180 = 144 p2 = 144+ 180 p2 = 324 p = √324 p = 18

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