Location of Roots when Compared with a constant 'k'
If both the r...
Question
If both the roots of the quadratic equation x2−6nx+(9n2−2n+2)=0 are greater than 3, then the range of n is:
A
(−∞,1)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
(−∞,119)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
[1,∞)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
(119,∞)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct option is D(119,∞) Given: x2−6nx+(9n2−2n+2)=0 Roots are greater than 3
To find: Range of n
Step-1: Draw the graph for the given expression y=x2−6nx+(9n2−2n+2)
Step-2: Write the applicable conditions.
Step-3: Solve all the conditions.
Step-4: Take intersection and combine the results for n.
Now, the applicable conditions that needs to be satisfied are: i).D≥0 ii).f(k)>0 iii).−b2a>k
Now, solving for all the conditions, i).D≥0 D=b2−4ac≥0⇒(−6n)2−4.1.(9n2−2n+2)≥0⇒36n2−36n2+8n−8≥0⇒8n−8≥0⇒8n≥8⇒n≥1⇒n∈[1,∞)⋯(A)
ii).f(k)>0, where k is 3 ⇒f(3)>0 ⇒32−6n×3+(9n2−2n+2)>0 ⇒9−18n+9n2−2n+2>0 ⇒9n2−20n+11>0 ⇒(9n−11)(n−1)>0 ⇒n<1 or n>119 ⇒n∈(−∞,1) or n∈(119,∞) ⇒n∈(−∞,1)∪(119,∞)⋯(B)
iii.)−b2a>k ⇒−6n2.1>3 ⇒3n>3⇒n>1 ⇒n∈(1,∞)⋯(C) ∴ Range of n, is A∩B∩C ⇒[1,∞)∩(−∞,1)∪(119,∞)∩(1,∞) ⇒A∩B∩C=(119,∞) ⇒n∈(119,∞)