Question

Find the value of k so that the following quadratic equation has equal roots:
2
$x {?}^{2}$
-(k-2) x + 1 = 0​

Answer 1

Answer:

${x}^{2} - (k - 2)x + 1 =0 \: has \: equal \: roots \\ so \: that \: d = 0 \\ d = {b}^{2} - 4ac \\ 0 = {b}^{2} - 4ac \\ 0 = { (- k + 2)}^{2} - 4(1)(1) \\ 0 = {k}^{2} - 4k + 4 - 4 \\ 0 = {k}^{2} - 4k \\ {k}^{2} = 4k \\ \bold \: {k = 0 \: and \: k = 4 }\\ hope \: it \: helps \: u...$

Answer 2

Step-by-step explanation:

For a quadratic equation,

$a {x}^{2} + bx + c$

Condition of equal roots is :-

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

Applying the formula,

${(k - 2)}^{2} - 4 = 0$

${k}^{2} - 4k = 0$

$k(k - 4) = 0$

$k = 0 \: or \: 4$

Regards

kshitij