Q. if the square of difference of the zeroe of the quadratic polynomial f(x)=x^2=px=45 is equal to 144, then find the value of p 

ANS.  x^2+px+45=0                                                                         let roots be a and b  hence, sum of root (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 and -18

in this answer how does 4ab comes

please answer fast

 

Asked by ketanjain.2502 | 14th May, 2019, 01:22: PM

Expert Answer:

In the solution we took (a - b) = (a + b)2 - 4ab.

R.H.S = (a + b)2 - 4ab

         = a2 + 2ab + b2 - 4ab

         = a2 -  2ab + b2 

         = (a - b)2 

         = L.H.S

Means (a - b) = (a + b)2 - 4ab is true.

So to get the solution we used the above identity.

Answered by Yasmeen Khan | 14th May, 2019, 02:07: PM