Question

find the values of K for which the quadratic equation K + 1 whole square + 2 ( K + 1) X + 1 is equal to zero​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Question:

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

Answer:

k = 0,-1

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 ;

(k+1)x² + 2(k+1)x + 1 = 0

Clearly , we have ;

a = k+1

b = 2(k+1)

c = 1

We know that ,

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

=> D = 0

=> [2(k+1)]² - 4•(k+1)•1 = 0

=> 4(k+1)² - 4(k+1) = 0

=> 4(k+1)(k+1-1) = 0

=> k(k+1) = 0

=> k = 0 , -1

Hence,

The required values of k are 0 and -1 .