Question

Find k 3xsquare +2kx+27=0 root are real & equal

Answer 1

Given equation : 3x^2 + 2kx + 27 = 0

On comparing the given equation with ax^2 + bx + c = 0 we get that a = 3 , b = 2k and c = 27.

We know, discriminant describes the nature of the root, and for equal & real roots discriminant is always equal to 0.

∴ b^2 - 4ac = 0

Substituting the values given in the question,

⇒ ( 2k )^2 - 4( 3 x 27 ) = 0

⇒ 4k^2 - 4( 81 ) = 0

⇒ 4{ k^2 - 81 } = 0

⇒ k^2 - 81 = 0

⇒ k^2 = 81

⇒ k = $\pm \sqrt{81}$

⇒ k = - 9 or 9.

Therefore the values of k satisfying the given information is 9 or - 9

Answer 2

P ( x ) = 3X² + 2KX + 27 = 0


Here,


a = Coefficient of X² = 3


B = Coefficient of X = 2K


And,


C = Constant term = 27.




The given equation have two real roots.


So,


Discriminant ( D ) = 0


B² - 4AC = 0

(2K)² - 4 × 3 × 27 = 0




4K² - 324 = 0



4K² = 324



K² = 91


K = √91


K = 9