Question

determine the value of k in the following equation(k-3)x2+x-6=0 to have real and equal roots​

Answer 1

Required value of k is

$k = \dfrac{71}{24}$

Step-by-step explanation:

We are given the following quadratic equation in the question:

$(k-3)x^2+x-6=0$

If the equation have real and equal roots, then the discriminant of the quadratic equation is zero.

General form of equation:

$ax^2 + bx + c = 0\\a = (k-3)\\b = 1\\c = -6$

Equating discriminant to zero.

$D = b^2 - 4ac = 0\\(1)^2 - 4(k-3)(-6) = 0\\24(k-3) = -1\\\\k-3 = \dfrac{-1}{24}\\\\k = \dfrac{-1}{24} + 3\\\\k = \dfrac{71}{24}$    

is the required value of k.  

#LearnMore

Determine the value of k for which the quadratic equation 2 X square + 3 X + K is equal to zero have equal and real roots

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Answer 2

The value of k is equal to  $\dfrac{71}{24}$.

Step-by-step explanation:

The given quadratic equation:

$(k-3)x^2$ + x - 6 = 0

Here, a = k - 3, b = 1 and c = - 6

To find, the value of k = ?

∴ D = $b^{2} -4ac$

= $1^{2}$ - 4( k - 3)(- 6)

= 1 + 24(k - 3)

= 1 + 24(k - 3)

= 1 + 24 k - 72

= 24 k - 71

∵ The roots are real and equal.

D = 0

∴ 24 k - 71 = 0

⇒ 24 k = 71

⇒ k = $\dfrac{71}{24}$

Thus, the value of k is equal to  $\dfrac{71}{24}$.