Question

Find the values of k for which the equation 2x^2 - kx + 2 = 0 has no real roots. Also can someone clearly define the no real roots part pls.
I got k<-4
Is that right?
As minus 4 is less than 0. Or is it completely wrong?

Answer 1

f(x) = 2x² - kx + 2

For no real roots,

Discriminant < 0

=> b² - 4ac < 0

=> (-k)² - 4(2) (2) < 0

=> k² - 16 < 0

=> k² < 16

=> k < √16

=> k < 4

Since, k < 4..i think it may contains negative as well as positive values

Answer 2

Answer:

Given that equation

$2 {x}^{2} - kx + 2 = 0$

has no real roots .

So, its discriminant

${b}^{2} - 4ac$

when will less than 0.

That is

$({ - k})^{2} - 4 \times 2 \times 2 < 0\\ \implies \: {k}^{2} - 16 < 0 \\ \implies(k + 4)(k - 4) < 0$

Final step is written using

${x}^{2} - {y}^{2} = (x + y)(x - y)$

Now we can find the value of k in set form by the help of real number line as shown in attachment .

After seeing the attachment we will get that

$k \in( - 4,4)$