the roots of x^2‐2x‐(r^2‐1)=0 are
Answers
We are given,
f(x)=x
2
−2x+(−a
2
+1)=0→(1)
here the co-efficient of x
2
is positive & g(x)=x
2
−(a+1)x+(a−1)=0→(2) & also here the co-efficient of x
2
is positive.
⇒ Let α
1
,β
1
are the roots of f(x) & α
2
,β
2
are the roots of g(x).
Now, here we can absorbed that at pt α
1
& β
1
g(x) will generated negative values.
So, g(α
1
),g(β
2
)<0
From equati0on (1)
α
1
or β
1
=
2a
−b±
b
2
−4ac
=
2(1)
−(−2)±
(−1)
2
−4(1)(a−1)
=
2
2±
4a
2
=1±a
So, α
1
=1+a & β
1
=1−a
as a∈R
Now, g(α
1
)=(1+a)
2
−(a+1)(1+a)+1(a−1)<0
1+2a+a
2
−a
2
−2a−1+a−1<0
a<1
→(1)
& g(β
1
)=(1−a)
2
−(a+1)(1−a)+1(a−1)<0
1−2a+a
2
−(a−a
2
+1−a)+a−1<0
1−2a+a
2
+a
2
−1+a−2<0
2a
2
−a−1<
2a
2
−2a+a−1<0
2a(a−1)+1(a−1)<0
(2a+1)(a−1)<0
→ at a>1 quantity >0, so condition False
→ now at a<−1/2 quantity >0, so condition again False
→ now at a ∈(−1/2,1) quantity <0
So, a∈(−1/2,1)→((1)
From (i) & (ii)
a∈(
2
−1
,1)
solution