Find the discriminant of the quadratic equation x² + kx - 3 = 0 in terms of k, where k ∈ R. Hence show that x² + kx - 3 = 0 has distinct real roots. Let α and β be those roots. Write down α + β in terms of k and the value of αβ . find the quadratic equation which has 2α + β and α + 2β as its roots.
Answers
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 −4k=0
9−4k=0
k=904
Therefore, p=3 and k= 9/4
- We have to find the discriminant of the quadratic equation x² + kx - 3 = 0 in terms of k, where k ∈ R.
- Show that x² + kx - 3 = 0 has distinct real roots.
- Let α and β be those roots. Write down α + β in terms of k and the value of αβ .
- find the quadratic equation which has 2α + β and α + 2β as its roots.
1. we know, D = b² - 4ac
= (k)² - 4(-3)(1)
= k² + 12
Therefore Discriminant in terms of k is k² + 12.
2. Here, Discriminant , D = k² + 12 > 0 for all real value of k.
hence roots of given quadratic are distinct and real.
3. sum of roots = α + β = - coefficient of x/coefficient of x²
⇒α + β = -(k)/1 = -k ...(1)
product of roots = αβ = constant/coefficient of x²
⇒αβ = (-3)/1 = -3 ...(2)
4. quadratic equations can be written as :
x² - (sum of roots)x + product of roots.
here, roots are 2α + β and α + 2β.
∴ sum of roots = (2α + β) + (α + 2β) = 3(α + β)
from equation (1) we get,
= 3(-k)
= -3k
product of roots = (2α + β)(α + 2β) = 2α² + 4αβ + αβ + 2β²
= 2(α² + β²) + 5αβ
= 2[(α + β)² - 2αβ] + 5αβ
= 2(α + β)² + αβ
from equations (1) and (2) we get,
= 2(-k)² + (-3)
= 2k² - 3
hence, equation ; x² - (2k² - 3)x + (-3k)
= x² - (2k² - 3)x - 3k