The sum of the roots of the equation x^2 - 4x = K(x - 1) - 5 is 7. Wha

For a standard quadratic equation in the form of a\(x^2\) + bx + c = 0, where a≠0, the sum of roots is given by -\(\frac{b}{a}\)
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In questions on Quadratic equations, it is always good practice to compare the given equation to the standard form and write down the values of a, b and c. This will always help you avoid silly mistakes, because you will be able to carefully consider the values of a, b and c with due respect to their signs.

First, let us rearrange the given equation to bring it to the form of a\(x^2\) + bx + c = 0

The given equation is \(x^2\) – 4x = K(x – 1) – 5

This can be re-written as \(x^2\) – 4x = Kx – K – 5.

Transferring all terms to the LHS, we have, \(x^2\) – 4x – Kx + K + 5 = 0.
Since K is a constant, K + 5 represents the constant term of this quadratic equation. Grouping the terms containing x, we have,

\(x^2\) – (4 + K)x + (K + 5) = 0.

Comparing with the standard form, a = 1, b = -(4+K) and c = (K+5).

We know that the sum of the roots of the given quadratic equation is 7.

Therefore, -\(\frac{b}{a}\) = 7

Substituting the values of b and a in the equation above, we have,

-{-\(\frac{(4 + K) }{ 1}\)} = 7

Simplifying, we have, 4 + K = 7 or K = 3.
The correct answer option is D.