Question

Find the values of k for which the quadratic equation (k + 4)x^2 + (k + 1)x + 1 = 0 has equal roots. Also find these roots.

Answer 1

⚽ Correct Question ⚽ $\:$

• Find the values of k for which the quadratic equation (k + 4)x^2 + (k + 1)x + 1 = 0 has equal roots. Also find these roots.

⚽ Solution ⚽

We know that, for a given equation ax²+bx+c=0 to have equal roots, the required condition is

> b²-4ac=0

⚽ Given, ⚽

(k+4)x²+(k+1)x +1=0 has equal roots

→(k+1)²-4(k+4)=0

→k²+1+2k-4k-16=0

→k²-2k-15=0

→k²-5k+3k-15=0

→k(k-5)+3(k-5)=0

→(k+3)(k-5)=0

→k=-3,5

If k=-3

⚽ Now, the polynomial becomes ⚽

→(k+4)x²+(k+1)x +1=0

→(-3+4)x²+(-3+1)x+1=0

→x²-2x+1=0

→(x-1)²=0

→x=1

If k=5

⚽ Now, the polynomial becomes ⚽

→(k+4)x²+(k+1)x +1=0

→9x²+6x+1=0

→9x²+3x+3x+1=0

→3x(3x+1)+1(3x+1)=0

→(3x+1)²=0

→3x+1=0

→x= -1/3

Therefore,

⚽ the roots are 1 (or)-1/3 ⚽

Answer 2

Answer

The given equation is

(k+1)x – 2(k – 1)x+1=0

comparing it with ax + bx +c = 0 we get

a = (k +1), b = -2(k – 1) and c=1 Since roots are real and equal, so = either k = 0ork – 3 = 0 = 4k (k –

.Discriminant, D = b – 4ac = 4(k – 1)2-4(k +1) x 1

= 4(k - 2k + 1) - 4k - 4

= 4k? - 8k + 4 - 4k - 4 = 4k2 - 12k

D = 0= 4k- 12k = 0 4k(k - 3) = 0

3) = 0

Hence, k = 0, 3.