Question

If (4-k)x²+2(k+2)x+(8k+1)=0 has equal roots then find the value of k. Plzz help me....

Answer 1

Since the roots here are equal, let both be a.

Now as we know that in a quadratic equation with coefficient of x = 1 , the product of thr roots give c and their sum gives b.

So by using that relation,

${x}^{2} + 2 \frac{(k + 2)}{(4 - k)}x + \frac{(8k + 1)}{(4 - k)} = 0$


So we have,
sum of roots

$a + a = 2 \frac{(k + 2)}{4 - k} \\ a = \frac{(k + 2)}{4 - k} \:$

And also,

product of roots,
${a}^{2} = \frac{8k + 1}{4 - k}$
using these two values

${( \frac{(k + 2)}{(4 - k)}) }^{2} = \frac{8k + 1}{4 - k} \\ \\ {k}^{2} + 4k + 4 = (4 - k)(8k + 1) \\ \\ {k}^{2} +4k + 4 = 32k + 4 - 8 {k}^{2} - k \\ 9 {k}^{2} - 27k = 0 \\ 9k(k - 3) = 0$

So we have k =0 and k= 3.


Tada!!