Question

If the equation x² – kx + 1, have no real roots, then *

Answer 1

Answer:

$\boxed{\bf \implies k< \pm2}$

Step-by-step explanation:

$Given, \\ the \ equation \ x^2-kx+1 \ has \ no \ real \ roots. \\\\ The \ equation \ is \ of \ the \ form \ ax^2+bx+c, \\ a=1,b=-k,c=1 \\\\ It \ has \ no \ real \ roots, means \\ the \ discriminant,D=b^2-4ac<0 \\\\ \implies (-k)^2-4(1)(1)<0 \\ \implies k^2-4<0 \\ \implies k^2<4 \\ \implies k<\sqrt{4} \\ \implies k< \pm2$

Answer 2

Solve when:

k is a real number.

If $\sf{x^2-kx+1}$ doesn't have any real roots, both roots must be imaginary.

Since roots are imaginary the discriminant must be negative.

The discriminant is $\sf{k^2-4<0}$.

After we solve the inequality we get $\sf{-2<k<2}$.

Solve when:

k is a complex. (outside the course)

There are more solutions other than $\sf{-2<k<2}$.

From the polynomial

• $\sf{\alpha +\beta =k}$ [Relation between Roots and Coefficients]
• $\alpha =\frac{1}{\beta }$ [Relation 〃]

What if $\sf{k=a+bi}$ ? [$\sf{i=\sqrt{-1} }$]

Since k is imaginary, $\alpha +\beta$ is also imaginary.

From this, we obtain two roots both purely imaginary.

Therefore the true solution is k ∈ {$\sf{-2<k<2}$ ∪ $\rm{\mathbb{C}-\mathbb{R}}$}.

• C is a complex set
• R is a real set
• So C-R is a purely imaginary set.