If the Squared difference of the zeros of the quadratic polynomial f(x)=x square + px + 45 is equal to 144 , find the value of "p".

Asked by Anubhav Sachan | 29th Apr, 2013, 08:33: PM

Expert Answer:

x^2 +px+45 =0
Let roots be a and b
hence, sum of roots (a+b) = p and product of roots (ab) = 45
 
Now, (a-b)^2  = 144
(a+b)^2 - 4ab = 144
p^2 - 4*45 = 144
p^2 = 324
p = +18 or -18

Answered by  | 30th Apr, 2013, 04:54: AM

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