Location of Roots when Compared with a constant 'k'
The value of ...
Question
The value of m for which atleast one root of the quadratic equation (m2+m+2)x2−(m+5)x+2=0 is greater than −1 is
A
[−1,−79]
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B
[−79,1]
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C
[−1,79]
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D
[79,1]
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Solution
The correct option is C[−1,79] Given: (m2+m+2)x2−(m+5)x+2=0;
have atleast one root is greater than −1.
let y=(m2+m+2)x2−(m+5)x+2
On comparing with standard form of quadratic equation y=ax2+bx+c we get a=m2+m+2,b=−(m+5),c=2. Taking a=m2+m+2=(m+12)2+32 ⇒a>0, hence we get an upward facing parabola
Now we will have 3 cases.
Case 1: Both the roots are greater than −1.
Required conditions are (i)D≥0 D=b2−4ac≥0 D=(−(m+5))2−4.(m2+m+2).(2)≥0 ⇒7m2−2m−9≤0 ⇒7m2+7m−9m−9≤0 ⇒7m(m+1)−9(m+1)≤0 ⇒(7m−9)(m+1)≤0 ⇒m∈[−1,79]⋯(1)
(iii)x coordinate of the vertex will be greater than −1 ∴−b2a>−1 −(−(m+5))2(m2+m+2)>−1 ⇒2m2+3m+9>0) ⇒m∈R⋯(3)
From equation (1),(2),(3) m∈[−1,79]⋯(4)
Case 2: One root is greater than −1 and other is less than −1.
Required condition is (i)f(−1)<0 (m2+m+2)(−1)2−(m+5)(−1)+2<0 ⇒m2+2m+9<0 ⇒m∈ϕ⋯(5)
Case 3: One root is greater than −1 and other root (smaller) is equal to −1.
Required conditions are (i)D≥0 D=b2−4ac≥0 D=(−(m+5))2−4.(m2+m+2).(2)≥0 ⇒7m2−2m−9≤0 ⇒7m2+7m−9m−9≤0 ⇒7m(m+1)−9(m+1)≤0 ⇒(7m−9)(m+1)≤0 ⇒m∈[−1,79]⋯(6)