Question

find that non zero value of k for quadratic equation kx2+1-2(k-1)x+x2=0has equal roots

Answer 1

Solution=

Given,

Kx2+1-2(k-1)x+x2

=x2(k+1)-2(k-1)+1

 

Here,

a=(k+1)

b=-2(k-1) & c=1

for real and equal roots,

d=0=b2-4ac=0

putting the values of a,b,c

[-2(k-1)]2-4(k+1)(1)=0

Then on opening the brackets we’ll get,

(4k)(k-3)=0

= 4k=0  and  (k-3)=0

= k=0 or k=3

Hope it helps

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Answer 2

Answer:

k = 3

Step-by-step explanation:

The quadratic equation is

$kx^2+1-2(k-1)x+x^2=0$

We can rewrite this equation as

$(k+1)x^2-2(k-1)x+1=0$

For the quadratic equation have equal zeros, the discriminant should be zero.

$b^2-4ac=0\\\\(-2(k-1))^2-4(k+1)=0\\\\4k^2-8k+4-4k-4=0\\\\4k^2 -12k=0\\\\4k(k-3)=0\\\\k=0,3$

Hence, the non zero value of k is 3.