Complete set of real values of so that equation y2 + (√k-1)x+K² -4 = 0 has one root less than 08 other greater than 0, is
Answers
Answer:
Correct option is
A
[
4
3
,∞)
Given, the equation x
4
−2ax+x+a
2
−a=0
Considering it a quadratic equation in a,
a
2
−a(1+2x
2
)+x
4
+x=0
Solving the equation we get,
a=
2
−[−(1+2x
2
)]±
[−(1+2x)]
2
−4(x
4
+x)
=
2
(1+2x
2
)±
1+4x
2
+4x
4
−4x
4
−4x
=
2
(1+2x
2
)±
4x
2
−4x+1
=
2
(1+2x
2
)±
(2x−1)
2
=
2
(1+2x
2
)±(2x−1)
∴a=
2
1+2x
2
+2x−1
,
2
1+2x
2
−2x+1
⇒a=x
2
+x,x
2
−x+1
Thus, we get a factorisation,
(x
2
+x−a)(x
2
−x+1−a)=0
For, x
2
+x−a=0
⇒x=
2
−1±
1+4a
For, the roots to be real 1+4a≥0
⇒a≥−
4
1
So, a has the roots in [−
4
1
,∞)
For, x
2
−x+1−a=0
⇒x=
2
−1±
4a−3
For, the roots to be real, 4a−3≥0
⇒a≥
4
3
So, a has the roots in [
4
3
,∞)
We see that the common interval is [
4
3
,∞)
Hence, the complete set of real values of a for which the given equation has all its real roots is [
4
3
,∞).