Question

[Maths]

$\large \underbrace{ \overbrace{ \over \underline{\texttt{Question:-}}}}$
Find the value of k for which the following equation has equal root.
$\dag \orange{ \tt \: \: {x}^{2} - 2kx + 7k - 12 = 0}$
Also, find the roots for those values of k.​

Answer 1

Answer:

Given equation,

→ x² - 2kx + 7k - 12 = 0

We have to find out the values of k for which the given equation has real and equal root.

Here,

→ a (coefficient of x²) = 1

→ b (coefficient of x) = -2k

→ c (constant term) = 7k - 12

Therefore, the value of the discriminant will be,

→ D = b² - 4ac ★

→ D = (-2k)² - 4 × (7k - 12)

→ D = 4k² - 28k + 48

Note that:

• If D > 0, roots are real and distinct.
• If D = 0, roots are equal.
• If D < 0, roots are imaginary.

So, if the given equation has equal roots,

→ D = 0

→ 4k² - 28k + 48 = 0

→ k² - 7k + 12 = 0

→ (k - 3)(k - 4) = 0

→ either (k - 3) = 0 or (k - 4) = 0

→ k = 3, 4

★ So, the possible values of k are 3 and 4 such that the roots of the given equation are equal.

Answer:

• k = 3, 4

•••♪

Answer 2

Given Equation

$\tt \to \: {x}^{2} - 2kx + (7k - 12) = 0$

To Find the value of x

Now Compare with

$\tt \to \: a{x}^{2} + bx + c = 0$

We get

$\tt \to \: a = 1,b = - 2k \: and \: c = 7k - 12$

According to question the equation has real root

$\tt \to \:D = 0$

Where

$\tt \to \: D = {b}^{2} - 4ac$

Put the value on formula

$\tt \to \: ( - 2k) {}^{2} -4 (7k - 12) \times 1 = 0$

$\tt \to \: 4 {k}^{2} -4(7k - 12) = 0$

$\tt \to 4\{ {k}^{2} - (7k - 12) \} = 0$

$\tt \to \: {k }^{2} - 7k + 12 = 0$

$\tt \to \: {k}^{2} - 4x - 3x + 12 = 0$

$\tt \to \: k(k - 4) - 3(x - 12) = 0$

$\tt \to \: (k - 3)(k - 4) = 0$

$\tt \to \: k = 3 \: and \: k = 4$

Answer

$\tt \to \: k = 3 \: and \: k = 4$