Show that the equation $$ x^{2}+k x=5-2 k $$ has two real and distinct roots for all real values of $$ k $$.

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We start with the equation

x² + k x = 5 − 2k.

First, we rewrite the equation in the standard quadratic form:

x² + k x − (5 − 2k) = 0.

For a quadratic equation ax² + bx + c = 0 to have two real and distinct roots, its discriminant (∆), given by ∆ = b² − 4ac, must be strictly positive.

In our equation, a = 1, b = k, and c = −(5 − 2k). Thus, the discriminant is

∆ = k² − 4(1)(−(5 − 2k))
    = k² + 4(5 − 2k)
    = k² + 20 − 8k.

It is often helpful to complete the square for such expressions. Notice that

k² − 8k + 20 = (k² − 8k + 16) + 4
           = (k − 4)² + 4.

Since (k − 4)² is always non-negative for all real k (as squares are always ≥ 0), it follows that

(k − 4)² + 4 ≥ 0 + 4 = 4.

In fact, more precisely, (k − 4)² + 4 > 0 for all real k because the minimum value occurs when (k − 4)² = 0, giving ∆ = 4.

Since ∆ = (k − 4)² + 4 is always greater than zero irrespective of the value of k, the quadratic equation has two real and distinct roots for all real values of k.
For all real values of $$ k $$, the equation $$ x^{2} + kx = 5 - 2k $$ has two real and distinct roots because the discriminant $$ \Delta = (k - 4)^2 + 4 $$ is always positive.

Show that the equation \( x^{2}+k x=5-2 k \) has two real and distinct roots for all real values of \( k \).

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