Identity Condition for Polynomial Equations

An equation of the form \(Ax^2 + Bx + C = 0\) is considered an identity if it holds true for all possible values of the variable \(x\). For this to be true, the coefficients of the polynomial must all be equal to zero. That is, we must have \(A = 0\), \(B = 0\), and \(C = 0\) simultaneously.

Setting Coefficients to Zero

In the given equation, \((k^2-5k+4)x^2+(k^2-3k-4)x+(k^2-4k)=0\), we identify the coefficients:

• \(A = k^2-5k+4\)
• \(B = k^2-3k-4\)
• \(C = k^2-4k\)

For the equation to be an identity, we set each coefficient to zero:

1. \(k^2-5k+4 = 0\)
2. \(k^2-3k-4 = 0\)
3. \(k^2-4k = 0\)

Solving for k in Each Equation

We solve each quadratic equation for \(k\) to find the possible values:

Equation 1: \(k^2-5k+4 = 0\)

Factoring the quadratic expression:

\((k-1)(k-4) = 0\)

This gives us two possible values for \(k\): \(k=1\) or \(k=4\).

Equation 2: \(k^2-3k-4 = 0\)

Factoring the quadratic expression:

\((k-4)(k+1) = 0\)

This gives us two possible values for \(k\): \(k=4\) or \(k=-1\).

Equation 3: \(k^2-4k = 0\)

Factoring the expression:

\(k(k-4) = 0\)

This gives us two possible values for \(k\): \(k=0\) or \(k=4\).

Finding the Common Value of k

For the original equation to be an identity, the value of \(k\) must satisfy all three conditions simultaneously. We look for the common value among the solutions from the three equations:

• Solutions for Eq 1: \(\{1, 4\}\)
• Solutions for Eq 2: \(\{-1, 4\}\)
• Solutions for Eq 3: \(\{0, 4\}\)

The only value of \(k\) that appears in all three sets of solutions is \(k=4\).

Conclusion

Therefore, the value of \(k\) for which the given equation is an identity is 4.