Explanation
The roots of (x−a)(x−c)+k(x−b)(x−d)=0 are real and distinct for all real k
f(x)=(1+k)x2−(a+c+k(b+d))x+ac+kbd
As D>0 for real roots.
So ,
⇒ (a+c+k(b+d))2−4(1+k)(ac+kbd)>0
⇒ (b+d)2k2+2(a+c)(b+d)k+(a+c)2−4(bdk2+(ac+bd)k+ac)>0
⇒ (b−d)2k2+(2(a+c)(b+d)−4(ac+bd))k+(a−c)2>0
This true for all k
D<0
(2(a+c)(b+d)−4(ac+bd))2−4((a−c)(b−d))2<0
((a+c)(b+d)−2(ac+bd))2−((a−c)(b−d))2<0
Applying a2−b2=(a−b)(a+b) we get,
⇒ ((a+c)(b+d)−2(ac+bd)+(a−c)(b−d))((a+c)(b+d)−2(ac+bd)−(a−c)(b−d))<0
⇒ (ab+ad+bc+cd−2ac−2bd+ab−ad−bc+cd)(ab+ad+bc+cd−2ac−2bd−ab+ad+bc−cd)<0
⇒ (ab+cd−ac−bd)(ad+bc−ac−bd)<0
⇒ (a−d)(b−c)(a−b)(d−c)<0
⇒ a<b<c<d
Given, rx2=1sx+3
⇒rx2−1sx−3=0
WE know that in equation ax2+bx+c=0
The roots of equation
x=−b±√b2−4ac2a
In given equation rx2−1sx−3=0
compare with above equation we get a=r, b=−1s and c=−3
Then x=−−1s±√(−1s)2−4×r×−32r
x=1s±√1s2+12r2r
⇒x=12sr±√1s2+12r2r
We have,
x2+px+q=0......(1)
x2+qx+p=0.......(2)
By equation (1)
Let a and b be the roots of equation (1)
Then,
Sum of roots a+b=−p1=−p
Product a.b=q1=q
According to given question,
Difference of roots
a−b=√(a+b)2−4ab
a−b=√(−p)2−4q
a−b=√p2−4q
Now, by equation (2)
Let c and d be the roots of equation (2)
Sum of roots c+d=−q1=−q
Product a.b=p1=p
c−d=√(c+d)2−4cd
c−d=√(−p)2−4q
c−d=√p2−4q
Again, according to given question,
Difference of roots of both equations are equal
Therefore,
a−b=c−d
√q2−4p=√p2−4q
q2−4p=p2−4q
q2−p2=4p−4q
(q−p)(q+p)=4(p−q)
(q−p)(p+q)=−4
(q−p)(p+q+4)=0
q=pandp+q+4=0
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