Question

find the value of k for which the quadratic equation 4 x square - 2 k + 1 X + K + 1 was equal roots
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Answer 1

Answer:

4x^2+x-k+1=0

Step-by-step explanation:

a=4,b=1,c=-k+1

D= b^2-4ac

=1+16k-16=0

k=15÷16

Answer 2

Question:

Find the value of k for which the quadratic equation 4x² + 2(k+1)x + k + 1 = 0 has equal roots.

Answer:

k = -1 , 3

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 ;

4x² + 2(k+1)x + k + 1 = 0

Clearly , we have ;

a = 4

b = 2(k+1)

c = k+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•4•(k+1) = 0

=> 4(k+1)² - 4•4(k+1) = 0

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

=> (k+1)(k-3) = 0

=> k = -1 , 3

Hence,

The required values of k are -1 and 3 .