CBSE Class 10 Answered
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 | 14 May, 2019, 01:22: PM
Expert Answer
In the solution we took (a - b)2 = (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)2 = (a + b)2 - 4ab is true.
So to get the solution we used the above identity.
Answered by Yasmeen Khan | 14 May, 2019, 02:07: PM
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