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Question

Find the set of values of m for which exactly one root of the equation
x2+mx+(m2+6m)=0 lie on (2,0)

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Solution

Let,

f(x)=x2+mx+(m2+6m)

As per question given that exactly one root of the equation lies between -2 and 0.


Hence, only one is negative.

Therefore f(2)×f(0)<0


Putting the value in the function ,we get

[(2)22m+m2+6m]×[m2+6m]<0

[m2+4m+4]×m(m+6)<0

(m+2)2×m×(m+6)<0


Hence, the range of m becomes

m(6,0){2,}

We can write it as

m(6,2)(2,0)


Hence, this is the answer .

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