Question

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

Answer 1

To find : the value of k for which the quadratic equation (k + 4)x² + (k + 1)x + 1 = 0 has equal

roots.

We know that :

Discriminant = b² - 4ac 

= (k+1)² - 4(1) × (k + 4)

= k² + 1 + 2k - 4k - 16 

                  

              = k² - 2k - 15

Given- roots are equal..

So,

k² - 2k - 15 = 0 

k² - (5 - 3)k - 15 = 0 

k² - 5k + 3k - 15 = 0 

k(k - 5) + 3(k - 5) = 0 

(k - 5)(k + 3) = 0 

k = 5 or k = -3 

2 values of k..

Take: k = 5 

(k + 4)x² + (k + 1)x + 1 = 0 

Putting k = 5,

(5 + 4)x² + (5 + 1)x + 1 = 0 

9x² + 6x + 1 =  0 

9x² + 3x + 3x + 1 = 0 

3x(3x + 1) + (3x + 1) = 0 

(3x + 1) (3x + 1) = 0 

(3x + 1)² = 0 

x = -1/3 

first root = -1/3

Now,

Take x = -3

(-3 + 4)x² + (-3 + 1)x + 1 = 0 

(1)x² + (-2)x + 1 =0 

x² - 2x + 1 = 0 

x² - x - x + 1 = 0 

x(x - 1) - (x - 1) = 0 

(x -1)(x - 1) = 0 

(x -1)² = 0 

x = 1 

Thus, other root = 1

So the Roots are 1 or -1/3 

Answer 2

Answer:

k = 5 , -3

Step-by-step explanation:

We Know

The condition for REAL & EQUAL roots is:

$b^{2}-4ac = 0$    (i)

Now

$p(x) = (k+4)x^{2} + (k+1)x +1$

Here

$a= (k+4) , b= (k+1) , c= 1$

Putting the values in eq. (i)

$(k+1)^{2} - 4(k+4)(1) = 0$

$k^{2} + 1 + 2k -4k-16 = 0 \\=k^{2} -2k -15 = \=k^{2} -2k=15$

$k^{2} -2k-15$

=$k^{2} +3k-5k-15$       [∵ on factorising middle term]

=$k(k+3) -5(k+3)$

=$(k-5)(k+3)$

⇒k-5=0║k+3=0

⇒k=5    ║k=-3

∴k = 5 , -3  Ans.

Hope it is helpful