X2 + kx +k = 0 has no real roots for x. (Q1) Find the discriminant of x2 + kx +k in terms of k?.?

Question

(Q2) Hence find the set of values that k can take?

x2 = x squared

Answer 1

OK. Compare you equation the the general form for a quadratic equation which is:

ax^2 + bx + c = 0

where a,b and c are constants and a does not = 0

You have a = 1, b = k and c = k

The discriminant for a quadratic is (b^2 - 4ac).

If the discriminant > 0 then there are two real roots

If the discriminant = 0 then there is one real root.

If the discriminant < 0 then there are no real roots.

Since your eqaution has nor real roots you know that that the discriminant b^2 - 4ac must be < than 0. Hence

k^2 - 4*1*k = 0

k^2 - 4k = 0

k(k - 4) = 0

So k = 0 or (k-4) = 0

Hence k = 0 or k = 4

PS Sorry, better complete it. Hence the set of values for k for which the given equation has no real roots is

0
Another way of expressing it would be:

[0,4]

Note, the square brackets mean 0 and 4 are included.

Answer 2

for a second degree equation to have no real roots, its graph does not cross the x axis. This happens when the discriminant is undefined. the discriminant would have been
square root of [k^2 - 4(1)k] which is the square root of a negative number

Therefore k^2 - 4k < 0
and so k(k - 4) < 0 which means that
k>0 when k - 4 < 0 or the interval 0 < k < 4

OR

k<0 when k - 4 >0 which has no solution

So the answer is ]0,4[ or using other notation 0 < k < 4

Answer 3

discriminant = b^2 - 4ac
for this equation
discriminant = k^2 - 4(1)(k) = k^2 - 4k
for no real roots, discriminant < 0
k^2 - 4k < 0

Q2

k^2 - 4k < 0
k(k-4)<0
k > 0 and k < 4

the answer is 0 < k < 4

Answer 4

a = 1, b = k c = k
the discriminant is k^2 - 4k = k(k-4)

for the eq to have no real roots.
k-4 < 0
0< k < 4