if A and B are roots of x2+px+q=0 and A4 and B4 are the roots of
x2-rx+s=0,then the equaton x2-4qx+2q2-r=0 ahs always :
1)2 real roots
2)2 +ve roots
3)2 -ve roots
4)one +ve and one -ve root
pleez answer soon
I got the answer as a) and d) both.
We have.
q = AB, r = A4+B4 and s = A4B4.
Now, D of given eqn. = 16q2-4(2q2-r)
= 8q2 + 4r
Now, q4=A4B4
=> q2=A2B2
Therefore, D = 4(A4+B4+2AB) = 4(A2+B2)2 >0
Now, roots of this eqn. are of the form:
[4q (+/-) sqrt(4(A2+B2)2)]/2
Now 2 cases arise
CASE I: q = +AB
Then roots are: [4AB (+/-) 2(A2+B2)]/2
= 2AB (+/-) A2+B2
= (A+B)2 and - (A-B)2
= +ve and -ve roots
CASE II: q = -AB
Then roots are: [-2AB (+/-) A2+B2]
= (A-B)2 and -(A+B)2
= +VE and -VE roots
Thus the given eqn. has 2 real roots out of which one is +ve and other is -ve.
We have.
q = AB, r = A4+B4 and s = A4B4.
Now, D of given eqn. = 16q2-4(2q2-r)
= 8q2 + 4r
Now, q4=A4B4
=> q2=A2B2
Therefore, D = 4(A4+B4+2AB) = 4(A2+B2)2 >0
Now, roots of this eqn. are of the form:
[4q (+/-) sqrt(4(A2+B2)2)]/2
Now 2 cases arise
CASE I: q = +AB
Then roots are: [4AB (+/-) 2(A2+B2)]/2
= 2AB (+/-) A2+B2
= (A+B)2 and - (A-B)2
= +ve and -ve roots
CASE II: q = -AB
Then roots are: [-2AB (+/-) A2+B2]
= (A-B)2 and -(A+B)2
= +VE and -VE roots
Thus the given eqn. has 2 real roots out of which one is +ve and other is -ve.
Sorry,
I made a mistake there. There will be no CASE II as q = + AB only.
The answer remains the same.
I made a mistake there. There will be no CASE II as q = + AB only.
The answer remains the same.
well
i got the same ans. but the ans given in the source is only (a)
Don't worry we both r right we know.
Have self-confidence now that we r right, after all we both got one common answer.
Have self-confidence now that we r right, after all we both got one common answer.