SOLUTION: I was givin the following question. The quadratic equation {{{ x^2+kx+k=0 }}} has no real roots for x. 1) Write down the discriminant of {{{ x^2+kx+k }}} in terms of k. 2)

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: I was givin the following question. The quadratic equation {{{ x^2+kx+k=0 }}} has no real roots for x. 1) Write down the discriminant of {{{ x^2+kx+k }}} in terms of k. 2)      Log On


   

Question 27170: I was givin the following question.
The quadratic equation +x%5E2%2Bkx%2Bk=0+ has no real roots for x.
1) Write down the discriminant of +x%5E2%2Bkx%2Bk+ in terms of k.
2) Hence find the set of values that k can take.
If somebody could help me. Thanks
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2+%2B+kx+%2B+k+=+0 has no real roots
that means the roots are imaginary- they involve sqrt%28-1%29, which is i.
The quadratic formula is:
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
where the equation is
a%2Ax%5E2+%2Bb%2Ax+%2B+c+=+0
but we have
x%5E2+%2B+k%2Ax+%2B+k+=+0
so,
a = 1
b = k
c = k
substitute these in the formula
x+=+%28-k+%2B-+sqrt%28+k%5E2-4%2A1%2Ak+%29%29%2F%282%2A1%29+
x+=+%28-k+%2B-+sqrt%28+k%5E2-4%2Ak+%29%29%2F2+
The discriminant, which is everything under the square root sign,
needs to be negative if the roots are imaginary
that means
+k%5E2+-+4%2Ak++%3C+0
factoring
+k%28k+-+4%29+%3C+0
k can't be greater than 4 or equal to 4.
that would make the inequality false
+4%284+-+4%29+%3C+0
4+%2A+0+%3C+0
not true
or
+5%285+-+4%29+%3C+0
5+%2A+1+%3C+0
not true either
k can't be zero or less than zero either
I would get
+0%280+-+4%29+%3C+0
not true
or
+-1%28-1+-+4%29+%3C+0
-1+%2A+-5+%3C+0
not true either
so, that means
4+%3E+k+%3E+0
meaning k is between 0 and 4 but doesn't include 0 or 4.