Question

Find the value of k for which quadratic equation
(3k+1)x^2+2(k+1)x+1 =0

Answer 1

(3k+1)x²+2(k+1)x +1=0

(3k+1)x²+(2k+2)x+ 1 =0

here comparing with standard form
we get

a=. 3k+1 b = 2k+2. c= 1


using real root


b²-4ac
(2k+2)². - 4(3k+1)

2k²+8k+4-12k-4

2k²-4k =0

2k²-2k-2k+0=0

2k(k-0) -0(2k -0)

k = 0

hence the value of k = 0

Answer 2

Question:

Find the value of k for which the quadratic equation (3k+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 ;

(3k+1)x² + 2(k+1)x + 1 = 0

Clearly , we have ;

a = 3k+2

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•(3k+1)•1 = 0

=> 4(k+1)² - 4(3k+1) = 0

=> 4[ (k+1)² - (3k+1) ] = 0

=> (k+1)² - 3k - 1 = 0

=> k² + 2k + 1 - 3k - 1 = 0

=> k² - k = 0

=> k(k-1) = 0

=> k = 0 , 1

Hence,

The required values of k are 0 and 1 .