Question

(vi) (3k+1)x^2+2(k+1)x+k=0 for which the roots are equal. Solve k​

Answer 1

Step-by-step explanation:

Given :-

(3k+1)x²+2(k+1)x+k=0

To find :-

(3k+1)x²+2(k+1)x+k=0 for which the roots are equal. Solve k ?

To find :-

Given Quadratic equation is (3k+1)x²+2(k+1)x+k=0

On Comparing this with the standard quadratic equation ax²+bx+c = 0

We have,

a = (3k+1)

b = 2(k+1)

c = k

If ax²+bx+c = 0 has equal roots then its discriminant D = b²-4ac = 0

On Substituting these values in the above formula then

=> [2(k+1)]² - 4(k+1)(k) = 0

=> 4(k+1)² - 4(k+1)(k) = 0

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

=> 4(k²+2k+1)-4(k²+k) = 0

=> 4k²+8k+4-4k²-4k = 0

=> (4k²-4k²)+(8k-4k)+4 = 0

=> 4k+4 = 0

=> 4k = -4

=> k = -4/4

=> k = -1

Therefore, k = -1

Answer :-

The value of k for the given problem is -1

Used formulae:-

• The standard quadratic equation ax²+bx+c = 0

• If ax²+bx+c = 0 has equal roots then its discriminant D = b²-4ac = 0

Points to know :-

• If D > 0 then it has two distinct and real roots.

• If D<0 then it has no real roots.