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Find the value of k for which the equation kx[x-2]+6=0 has equal roots.

@Himanshu4: Good answer! Keep posting!

 

The given equation is kx (x – 2) + 6 = 0.

kx (x – 2) + 6 = 0

∴ kx2 – 2kx + 6 = 0

Here a = k, b = – 2k and c = 6.

Discriminant of the quadratic equation = D = b2 – 4ac = (–2k)2 – 4 × k × 6 = 4k2 – 24k

Roots of the given quadratic equation are equal.

∴ D = 0

⇒ 4k2 – 24k = 0

⇒ 4k(k – 6) = 0

⇒ k = 0 or k – 6 = 0

⇒ k = 0 or k = 6

∴k = 6           (∵ k ≠ 0)

Thus, the value of k is 6.

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Problem = kx2 - 2kx + 6 = 0

Here a = k, b = 2k and c = 6

If roots are equal, then b2 - 4ac = 0

So, (2k)2 - 4(k)(6) = 0

=> 4k2 - 24k = 0

=> 4k(k - 6) = 0

=> k = 0 or k = 6

HOPE IT HELPS...

THUMBS UP PLZ...

  • 39

Kx2-2kx+6=0

a=K,b=-2k,c=6

D=b2- 4ac

D= (-2k)2-4*k*6

=4k2-24k

=4k(k-6)
If Quad. equ. has equal roots then D=0

4k(k-6)=0

(k-6)=0

K=6

  • 11

kx [ x-2 ]+6=0    

=>kx2 -2kx+6=0 has equal roots

therefore, D=0 i.e. b2-4ac=0

=>4k2 -24k=0

=>4k(k-6)=0

=>k=0 or k=6

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Thank u guyzz...

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