1.
If a,b are the roots of x2=x+1 , prove that a5=5a+3
2.
Two candidates attempt to solve a quadratic eqn of the form x2+px+q=0 . One starts with a wrong value of p and finds the roots to be 2 and 6. The other starts with a wrong value of q and finds the roots to be 2 and -9. Find the correct roots and the eqn.
392.Starting with the wrong value of p, lekin q ka sahi value hai na?
Putting incorrect roots (due to p) in eq, we get -:
2p + q = -4
6p + q = -36
giving -4p = 32, or p = -8 for which q = -4 + 16 = 12.
Putting incorrect roots(due to q) in eq, we get -:
2p + q = -4
-9p + q = -81
giving 11p = 77 or p = 7, q ka value yahan galat hai toh nahi chaiye.
Using p = 7 and q = 12, we get the equation
x2 + 7x + 12 = 0 whose roots are -3 and -4.
Dunno if I've solved this right. Just took a guess.
23yup pritish,
in first case value of p is wrong but q is correct
hence product of roots = 2 x 6 =12
in second case value of p is correct
hence sum of roots = -9+2= - 7
hence roots are -3 ,- 4
231st ques to easy hai
since a is a root ,
a2= a+ 1 ......(1)
thus a = a2- 1
(a2-1)2 = a + 1
a4 - 2a2 + 1 = a + 1
a3 - 2a = 1
a3 = 2a + 1 ..........(2)
multiply 1 and 2
a3a2 = (a+1)(2a+1) = 2a2+ 3a + 1
a5= 2a2+ 1 + 3a
= 2( a+ 1 ) + 1 + 3a .......(frm 1 )
= 5a+ 3
