What is the value of m for which #(4m+1)x^2 - 6mx+4# is a perfect square?

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Answer 1 · 2018-01-24T11:09:52

# m=(2(1+-sqrt10))/9#.

We know that, the Quadratic poly. #ax^2+bx+c# is perfect square

#iff Delta=b^2-4ac=0#.

In our case, #a=(4m+1), b=-6m, and, c=4#.

#:. Delta=0 rArr (-6m)^2-4(4m+1)(4)=0#,

#rArr 36m^2-16m-16=0#

#rArr 9m^2-4m-4=0............"[dividing the eqn. by "4]#,

Applying the Quadr. Formua, we get,

#m={-(-4)+-sqrt((-4)^2-4*9(-4))}/(2*9), i.e., #

# m=(4+-sqrt160)/18=(4+-4sqrt10)/18=(2(1+-sqrt10))/9#.