Question

for what value of k, the given expression is a perfect square
k^2x^2+2(k+1)x+4

Answer 1

please answer fast....its urgent...

Answer 2

hint: use disc to solve it

Answer 3

how??? 2nd term is not 4kx

Answer 4

and u r supposed to find value of k

Answer 5

ah yes I forgot the +1.

Answer 6

you can solve for k by using the quadratic formula

Answer 7

Determinant of quadratic must be equal to zero.

Answer 8

determinant or discriminant???
if u r talking about discriminant then plz explain why discriminant is zero in this case......bcz according to my knowledge disc=0 iff roots are equal......plz explain if u can..........

Answer 9

oh sorry discriminant :) :D

Answer 10

roots are equal \(x_1=x_2=\frac{b}{a}\)\[k^2x^2+2(k+1)x+4=(ax-b)(ax-b)=(ax-b)^2\]which is a perfect square :)

Answer 11

If it's a perfect square, the constant term of the binomial must be \(\sqrt{4} = 2\). Then our binomial is \((\sqrt{k^2x^2}+\sqrt{4}) = (kx+2)\) and when squared and expanded \[(kx+2)^2 = k^2x^2+4kx+4\]but we also have\[k^2x^2+2(k+1)x + 4\]so that means\[2(k+1)x = 4kx\]Discarding common factors\[k+1=2k\]\[k=1\]

Solving via discriminant: discriminant = 0 if perfect square
\[a = k^2\]\[
b=2(k+1)\]\[
c=4\]
Discriminant \( = b^2-4ac = 0 = 2^2(k+1)^2-4*4*k^2\)\[(k+1)^2-4k^2=0\]\[k^2+2k+1-4k^2=0\]\[-3k^2+2k+1=0\]\[(3k+1)(1-k) = 0\]\[k=1,k=-1/3\]However \(k=-1/3\) is extraneous because we don't get a perfect square with it. Demonstration left for the reader :-)