Question

(4- k) x square +(2k+4)x + 8k+ 1 = 0. Find the value of k​

Answer 1

The given quadric equation is (4 - k)x² + (2k + 4)x+ (8k+ 1) = 0, and roots are real and equal

Then find the value of k.

Here, a = (4-k), b = (2k + 4) and c = (8k

+ 1)

As we know that D = b² - 4ac

Putting the value of a = (4 - k), b = (2k + 4) and c = (8k + 1)

= (2k + 4)² - 4x (4 - k) x (8k+ 1)

= 4k² + 16k-16-4(4 + 31k-8k²)

= 4k² + 16k - 16 - 16 - 124k + 32k²

= 36k² - 108k+0

= 36k²-108k

The given equation will have real and equal roots, if D = 0

Thus,

36k²-108k = 0

18k(2k-6) = 0

k(2k-6) = 0

Now factorizing of the above equation

k(2k-6) = 0

So, either

k = 0

or

2k-6 = 0

2k = 6

k = 6/2 = 3

Therefore, the value of k = 0, 3.

Answer 2

$\footnotesize\tt{a = 4 - k}$

$\footnotesize\tt{b = 2k + 4}$

$\footnotesize\tt{c= 8k + 1}$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\boxed{ \underline{ \underline{\footnotesize \tt{❥ \: Given:- }}}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\footnotesize\tt{D = 0}$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\boxed{ \underline{ \underline{\footnotesize \tt{❥ \: Solution:- }}}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\footnotesize\tt{ {b}^{2} - 4ac = 0}$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\footnotesize \tt{ {(2k + 4)}^{2} - 4 \times (4 - k)(8k + 1) = 0}$

$\footnotesize \tt{ {4k }^{2} + 16k + 16 - 4(32k - {8k}^{2} + 4 - k) = 0}$

$\footnotesize \tt{ {4k }^{2} + 16k + 16 - 128k + {32k}^{2} - 16 + 4k = 0}$

$\footnotesize \tt{ {4k }^{2} + 16k + \cancel{16} - 128k + {32k}^{2} - \cancel{16} + 4k = 0}$

$\footnotesize \tt{ {4k }^{2} + {32k}^{2} - 128k + 16k + 4k = 0}$

$\footnotesize \tt{ {36k}^{2} - 108k = 0}$

$\footnotesize \tt{ 36k(k- 3) = 0}$

$\footnotesize \tt{ 36k = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: | \: \: \: \: \: \: \ \: \: \: \: \: \: k - 3 = 0 }$

$\footnotesize \tt{ k = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: | \: \: \: \: \: \: \ \: \: \: \: \: \: \: \: \: \: \: k = 3 }$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\boxed{ \underline{ \underline{\footnotesize \tt{❥ \: k = 0 }}}}$

$\boxed{ \underline{ \underline{\footnotesize \tt{❥ \: k = 3}}}}$