Question

1. Find the values of k such that the quadratic equation x2 - 2kx + (7k - 12) = 0
has equal roots.​

Answer 1

We have:-

• Quadratic equation = x² - 2kx + (7k - 12) = 0
• It have equal roots.

We have to FinD:-

• The value of k for which the condition is satisfied?

Solution:-

The given quadratic equation have real and equal roots. First of all, comparing with ax² + bx + c.

• a = 1
• b = -2k
• c = 7k - 12

We know that, D (Discriminate) = b² - 4ac and for real and equal roots, D = 0. Then,

➝ (-2k)² - 4(1)(7k - 12) = 0

➝ 4k² - 28k + 48 = 0

➝ k² - 7k + 12 = 0

Now finding the values of k by middle term factorisation or spilting the middle term method,

➝ k² - 3k - 4k + 12 = 0

➝ k(k - 3) - 4(k - 3) = 0

➝ (k - 4)(k - 3) = 0

Equating to zero, we will get k = 3,4

Hence:-

• The required unknown values of k possible is 3 or 4.

Answer 2

Step-by-step explanation:

Given :

• the quadratic equation x2 - 2kx + (7k - 12) = 0 has equal roots.

To Find :

• Find the values of k

Solution :

• According to the Ax² + Bx + c = 0

Then,

• A = 1

• b = -2k

• c = 7k - 12

• According to the d = b² - 4ac

Substitute all values :

D = 4k² - 4 × 1 × (7k - 12)

D = 4k² - 4 ( 7k -12 )

D = 4k² - 28k + 48

D = k² - 7k + 12

D = k² - 4k - 3k + 12

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

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

k = 4 , k = 3.

• Hence the k is 4 and 3