graph of quadratic equation when one root is greater than x1 and other is less than x2
Answers
Correct Question : Let x2-(m-3)x+m=0(mεR) be a quadratic equation . Find the values of m for which the roots are (ix)one root is smaller than 2 & other root is greater than 2 (x) both the roots are greater than 2 (xi) both the roots are smaller than 2 (xii)exactly one root lies in the interval (1,2) (xiii) both the roots lies in the interval (1,2) (xiv) atleast one root lies in the interval (1,2) (xv) one root is greater than 2 and the other root is smaller than
(x) one root is < 2 and other is > 2
From graph we get, D≥0
m∈(-∞,1)∪[9,∞) (1)
a*f(2)=1*f(2)<0
4-(m-3)⋅2+m<0
m+4-2⋅m+6<0
m>10 (2)
take the intersection of interval (1)
Ans=m∈(10,∞)
Again from the graph
D≥0
m∈(-∞,1)∪[9,∞) (1)
a⋅f(2)>0 f(2)=10-m>0 m<10
(2) Now vertex
-b2⋅a≥2
m-3>4
take intersection of intervals (1) and (2) Ans=m∈[9,10]
(xi) boot root <2 from graph 1) D≥0
m∈(-∞,1)∪[9,∞) (1)
2)a⋅f(x)>2 ⇒m<10 (2)
3)-b2a<2
m-33<2 (3) take the intersection of inervals (1),(2),(3) Ans=
m∈(-∞,1] xii) Exactly one root in interval (1,2)
D≥0
m∈(-∞,1)∪[9,∞) (1)
f(1)⋅f(2)<0 (1-m+3+m)⋅(4-2m+6+m)<0
4⋅(10-m)<0
m∈[10,∞)=Ans xiii)
Boot roots Lies (1,2) from posibilities by graph
D≥0
m∈(-∞,1)∪[9,∞) (1)
f(1)>0 and f(2)>0
4>0
1<-b2⋅a<2
1<m-22<2
2<(m-3)<4
5<m<7
now taking intersection from both the intervals
no such common intervals or no such m exist xiv)
both the roots can not lie in the interval
⇒ 1 root lies ∈ (1,2)
D≥0
m∈(-∞,1)∪[9,∞) (1)
f(1)⋅f(2)<0
4⋅(10-m)<0
intersection m∈(10,∞) xv)
one root >2
or the root <1
D≥0
m∈(-∞,1)∪[9,∞) (1)
f(1)<0 f(2)<0
since,f(1)=4(+ive)
Ans=No such m exists.
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One root greater than X1 and other root less than X2 can be also written as - both roots between X1, X2.
The graph of the above situation is in the attachment :-
Hope it helps you:)
