Find the discriminant of the quadratic equation x² + kx - 3 = 0 in terms of k, where k ∈ R. Let α and β be those roots. find the quadratic equation which has 2α + β and α + 2β as its roots.
Answers
Answer:
First suppose that the roots of the equation
x2−bx+c=0(1)
are real and positive. From the quadratic formula, we see that the roots of (1) are of the form
b±b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√2.
For the root or roots to be real, we require that b2−4c≥0, that is, b2≥4c. In order for them to be positive, we require that
b−b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√>0.
This immediately tells us that b>0, but we can go further. We can rearrange this to get
b>b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√,
which (assuming that b>0) is true if and only if
b2>b2−4c,
since both sides of the inequality are positive so we may square. But then
4c>0.
That is, if the roots are real and positive then b>0 and b2≥4c>0.
Now suppose that b>0 and b2≥4c>0.
Then the roots of (1) are real since b2−4c≥0, and b>0 guarantees that the root b+b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√ is positive.
So it remains to show that b−b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√>0. We have that
4c>0,
so that
b2>b2−4c,
then square rooting shows that
b>b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√,
so the roots of (1) are real and positive, as required.
Approach 2
It is given that -4 is a root of the equation x
2
+px−4=0.
Then, -4 must satisfy the equation,
(−4) ²+p×(−4)−4=0
16−4p−4=0
4p=12
p=3
Now, the equation x ²
+px+k=0 has equal roots
Therefore,
Discriminant, D=0⟹b ²
−4ac=p 2
−4k=0
9−4k=0
k= 9/4
Therefore, p=3 and k= 9/4