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.