For What Value of k is the Given Quadratic Equation a Perfect Square?

Question:

\( (4-k)x^2+(2k+4)x+(8k+1)=0 \)

Find the value of \(k\) for which the quadratic equation is a perfect square.

Solution

A quadratic expression is a perfect square when its discriminant is zero.

\( D=b^2-4ac=0 \)

Here,

\( a=4-k,\quad b=2k+4,\quad c=8k+1 \)

Substituting in the discriminant formula,

\( (2k+4)^2-4(4-k)(8k+1)=0 \)

\( 4k^2+16k+16-(124k+16-32k^2)=0 \)

\( 36k^2-108k=0 \)

\( 36k(k-3)=0 \)

\( k=0 \quad \text{or} \quad k=3 \)

Answer

\( \boxed{k=0 \text{ or } k=3} \)