Question

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

Answer 1

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 .

Answer 2

$\huge\mathfrak\blue{Answer:-}$

For real and equal roots,

D = 0

b² - 4ac = 0

Here,

a = 4

b = -2(k + 1)

c = (k + 1)

=>  -2²(k + 1)² - 4 x 4 x (k + 1) = 0

=> 4(k² + 2k + 1) - 16(k + 1) = 0

=>4k² + 8k + 4 - 16k - 16 = 0

=> 4k² - 8k - 12 = 0

=> k² - 2k - 3 = 0

=> k² -3k + k - 3 = 0

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

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

k = -1 or k = 3

∴ The values of k are -1 and 3