Question

11 Prove that


`{x}^(2) + kx + 2 = 0`
has real roots if

`k \geqslant 2 √(2)`
For which other values of k does the equation have real roots?​

Answer 1

9514 1404 393

Answer:

• the discriminant is non-negative for k ≥ 2√2
• k ≤ -2√2 will also give real roots

Explanation:

The discriminant of this quadratic equation is ...

d = b² -4ac

d = k² -4(1)(2) = k² -8

For real roots, we require ...

k² -8 ≥ 0

k² ≥ 8

|k| ≥ √8 = 2√2

That is, the equation will have real roots when ...

k ≥ 2√2 . . . . . . . what you're asked to show

k ≤ -2√2 . . . . . alternate values of k