Math, asked by NafePathan, 1 year ago

find non zero value of k, for which the quadratic equation kx²+1 - 2(k-x)+x²=0

Answers

Answered by Anonymous
10
Hi there!

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,

b² - 4ac = 0

[-2(k-1) ]²- 4(k+1)(1) = 0

(4k) (k-3) = 0

k = 3 or k = 0 [ Rejected. as Iqes. need non-zero value of k ]

Hence,
Non-Zero value of k is 3

Hope it helps!
Answered by Anonymous
3

Question:

Find the non zero value of k for which the quadratic equation kx² + 1 - 2(k-x) + x² = 0 has equal roots.

Answer:

k = -1/2

Note:

• An equation of degree 2 is know as quadratic equation .

• Roots of an equation is defined as the possible values of the unknown (variable) for which the equation is satisfied.

• The maximum number of roots of an equation will be equal to its degree.

• A quadratic equation has atmost two roots.

• The general form of a quadratic equation is given as , ax² + bx + c = 0 .

• The discriminant of the quadratic equation is given as , D = b² - 4ac .

• If D = 0 , then the quadratic equation would have real and equal roots .

• If D > 0 , then the quadratic equation would have real and distinct roots .

• If D < 0 , then the quadratic equation would have imaginary roots .

Solution:

The given quadratic equation is ;

kx² + 1 - 2(k-x) + x² = 0

=> kx² + 1 - 2k + 2x + x² = 0

=> (k+1)x² + 2x + 1 - 2k = 0

Clearly , we have ;

a = k + 1

b = 2

c = 1 - 2k

We know that ,

The quadratic equation will have equal roots if its discriminant is equal to zero .

=> D = 0

=> 2² - 4•(k+1)•(1-2k) = 0

=> 4 - 4•(k+1)(1-2k) = 0

=> 4•[ 1 - (k+1)•(1-2k) ] = 0

=> 1 - (k - 2k² + 1 - 2k) = 0

=> 1 - (-2k² - k + 1) = 0

=> 1 + 2k² + k - 1 = 0

=> 2k² + k = 0

=> k(2k + 1) = 0

=> k = 0 , -1/2

=> k = -1/2 (appreciate value)

Hence,

The required values of k is -1/2 .

Similar questions