Prove that the roots of the equation $(x-a)(x-b)=k^2$ are real for all values of k.

Question

Prove that the roots of the equation $(x-a)(x-b)=k^2$ are real for all values of $k^2$.

My Attempt:

$$(x-a)(x-b)=k^2$$ $$x^2-bx-ax+ab-k^2=0$$ $$x^2-(a+b)x+(ab-k^2)=0$$

Then, Discriminant $D=[-(a+b)]^2 - 4 \cdot 1 \cdot (ab-k^2)$ $$=(a-b)^2+4k^2$$.

Answer 1

You just need a finishing line assuming $a,b,k \in \mathbb{R}$.

Since $(a-b)^2+4k^2 \geq 0$, that is the discriminant is nonnnegative, the roots are real for all $k$.

Answer 2

Observing graphs $y=(x-a)(x-b)$ and $y=k^2$ you can say even more about both roots.

Answer 3

Without loss of generality let $a \ge b$.

Then take $x=|k|+a$.

We get, $(x-a)(x-b)=(|k|) \cdot (|k|+a-b)$ and since $a-b \ge 0$ hence we get,

The product $(x-a)(x-b) \ge k^2$

Now, note that if $x=a$ then $(x-a)(x-b)=0 \le k^2$

And hence we now have a value of $x$ for which $(x-a)(x-b) \ge k^2$ and a value of $x$ for which $(x-a)(x-b) \le k^2$.

Thus, by Intermediate Value Property

There mus exist some real number value of $x$ for which $(x-a)(x-b) = k^2$.

Thus the equation $(x-a)(x-b)-k^2=0$ has at least one real root.

Further, we know that complex non-real roots must occur in pairs and hence, the other root cannot be a complex non-real root.

Thus, given any $k \in \mathbb{R}$

The roots of the equation $(x-a)(x-b)=k^2$ must always be real numbers.

--Hence Proved--