The value of t for which both roots of equation x^2– 2(t+1)x+ 2 = 0 is positive :
Answers
Answer:K
=
−
5
Explanation:
A quadratic equation basic formula is
a
x
2
+
b
x
+
c
so matching up with
x
2
+
2
(
K
+
1
)
x
+
k
+
5
=
0
gives
x
2
terms:
x
2
∴
a
=
1
x
terms:
2
(
K
+
1
)
or
2
K
+
2
∴
b
=
2
K
+
2
constants:
K
+
5
∴
c
=
K
+
5
For a quadratic to have at least one positive root, the discriminate
(
Δ
)
must be equal to (1 real root) or greater than (two real roots) 0.
From the quadratic formula, the discriminate is
b
2
−
4
a
c
. We need
b
2
≥
4
a
c
So in this case,
(
2
K
+
2
)
2
≥
(
4
⋅
1
⋅
(
K
+
5
)
)
∴
4
K
2
+
4
K
+
4
≥
4
K
+
20
K
2
+
K
+
1
≥
K
+
5
K
2
+
1
≥
+
5
K
≥
4
So if K=4 then
x
2
+
2
(
4
+
1
)
x
+
(
5
+
4
)
=
0
x
2
+
10
x
+
9
=
0
(
x
+
9
)
(
x
+
1
)
∴
x
=
−
9
and
x
=
−
1
These are both negative so
K
≠
4
It seems like we need the c part to cancel. So if
K
=
−
5
then
the equation would be
x
2
+
2
(
−
5
+
1
)
x
+
(
5
−
5
)
=
0
x
2
−
8
x
+
0
=
0
x
(
x
−
8
)
∴
x
=
0
and
x
=
8
Step-by-step explanation:
f(x)=x
2
−2(a−1)x+2a+1
boththerootsaregreaterthanzero
∴f(0)>0
f(0)=0−0+2a+1>0
a+
2
1
>0
a>
2
−1
2
2(a−1)
>0
a−1>0
a>1
D≥0
4(a−1)
2
−4(2a+1)≥0
a
2
−2a+1−2a−1≥a
a
2
−4a≥0
a(a−4)≥0
a∈(−∞,0)U(4,∞)
∴a∈(4,∞)
soleastintergralvalueofais4