Question

find the value of k for a quadratic equation k X square +(2k+4)x+9=0​

Answer 1

Answer:         K=1                     or        K=4

Step-by-step explanation:

CORRECT EQUATION:

$kx^{2}+(2k+4)x+9=0$

Now, for equal and real roots,

$b^{2}-4ac=0$

Plugging in the values of a, b and c

we get,

$(2k+4)^{2}-4(k)(9)=0\\=>4k^{2}+16+16k-36k=0\\\\=>4k^{2}-20k+16=0\\=>4(k^{2}-5k+4)=0\\\\=>k^{2}-5k+4=0\\=>k^{2}-4k-1k+4=0\\=>k(k-4)-1(k-4)=0\\\\=>(k-1)(k-4)=0\\=>k=1\;\;\;or\;\;\;k=4\\\\\\Learning\;\;together\;:)$