Real and distant roots?

Question

Find the set of values of the constant k such that the equation x^2-6x+k=0 has real and distinct roots.



Yeah so what???

Answer 1

What you have to do is to remember the quadratic formula.

x = [-b +/- sqrt(b^2 - 4ac)]/[2a]

The only way there can be real and distinct roots is if the discrimant, b^2 - 4ac (i.e. the stuff inside the square root of the quadratic formula) is strictly greater than 0. If it was equal to 0, then there would be no distinct roots (since "plus or minus" 0 yields the same solution), and if it were less than 0, that would make the stuff inside the square root negative, and you can't take the square root of a negative number in the real numbers realm.

So, we calculate the inequality b^2 - 4ac > 0

(-6)^2 - 4(1)(k) > 0
36 - 4k > 0
-4k > -36

Remember that when we multiply or divide an inequality by a negative number, the greater sign flips to a less than.

k < 9

Therefore, the set of values are real numbers k such that k < 9.

Answer 2

I'm sure you mean DISTINCT roots. :)

Put the k on the other side:

x² -6x = -k

Complete the square:

x² -6x + 9 = -k + 9
(x - 3)² = 9-k
x - 3 = ±√(9-k)
x = 3 ±√(9-k)

So x will have real and DISTINCT roots for any k such that the value under the square root is greater than 0. (If it weren't for the "distinct", it would be >= 0.)

9 -k > 0
k < 9

Answer 3

To have real and distinct roots:
Delta > 0

Delta = b²-4ac

Delta = 6² - 4*1*k
=>
36 - 4k > 0
4k < 36
k < 9

₢

Answer 4

delta>0
36-4k>0
36>4k
9>k