Question

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

Answer 1

for a quadratic equation to have equal roots,
b^2 - 4ac = 0
here a = k+4, b=k+1, c=1
so
(k^2 + 2k +1) - 4(k+4)1 = 0
k^2 - 2k - 15 = 0
k^2 - 5k + 3k - 15 = 0
k(k-5) + 3(k-5) = 0
k=-3,5

Answer 2

Question:

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

Answer:

k = -3 , 5

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+4)x² + (k+1)x + 1 = 0

Clearly , we have ;

a = k+4

b = k+1

c = 1

We know that ,

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

=> D = 0

=> (k+1)² - 4•(k+4)•1 = 0

=> (k+1)² - 4(k+4) = 0

=> k² + 2k + 1 - 4k - 16 = 0

=> k² - 2k - 15 = 0

=> k² - 5k + 3k - 15 = 0

=> k(k-5) + 3(k-5) = 0

=> (k-5)(k+3) = 0

=> k = -3 , 5

Hence,

The required values of k are -3 and 5 .