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Question

x2(m3)x+m=0(mR) be a quadratic equation. Find the value of m for which both roots are real and distinct:

A
(1,1)
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B
(,1)U(9,)
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C
(,4)U(12,)
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D
(4,3)U(5,)
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Solution

The correct option is B (,1)U(9,)
Given quadratic equation is x2(m3)x+m=0.

Also given that both roots are real and distinct.

We have to find the value of m.

Since the roots are real and distinct, the discriminant b24ac is greater than 0.

We have a=1,b=(m3),c=m

We have b24ac>0

((m3))24m>0

(m3)24m>0

m26m+94m>0

m210m+9>0

(m9)(m1)>0

Computing signs of each term we get

m9=0m=9,m9<0m<9,m9>0m>9

m1=0m=1,m1<0m<1,m1>0m>1

Summarize the signs in a table we get
m<1m=1 1<m<9m=9 m>9
m9 0 +
m1 0 + + +
(m9)(m1) + 0 0 +
Thus (m9)(m1)>0 when m<1 and m>9 .

Therefore the value of m is (,1)(9,).

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