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

1 Answer
Jan 24, 2018

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

Explanation:

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.