Question

the value of k for which the equation x2 - 4x +k = 0 has two distinct real roots is

Answer 1

━━━━━━━━━━━━

$\huge\bold\star\red{answer:-}$

For a quadratic equation 

ax²+bx+c=0 to have equal roots, D=b2−4ac=0

∴D=b2−4ac=(−4)2−4(1)(k)=0

∴k=4

Answer 2

$\large \underline \mathfrak{solution}$

${ \rm {x}^{2} - 4x + k > 0}$

${ \rm{here}}$

${ \rm{ {ax}^{2} + bx + c = 0}}$

${ \rm{so \: a = 1 \: b = 4 \: c = k}}$

${ \rm{ \therefore {b}^{2} - 4ac > 0}}$

${ \rm{ substituing \: the \: value}}$

${ \rm{ \to {4}^{2} - 4.1.k > 0}}$

${ \rm{ \to 16 - 4k > 0}}$

${ \rm{ \to 16 > 4k }}$

${ \rm{ \to \frac{16}{4} > k}}$

${ \rm{ \to 4 > k}}$

${ \rm{ \to k < 4}}$

${ \rm{ \large{so \: the \: value \: is \: k < 4}}}$

──────────────────────────

If for a quadratic eqation ${ \rm{ {ax}^{2} + bx + c = 0}}$ where a, b, c are real numbers and a≠0, then discriminat :

(i) [tex]{ \rm{ {b}^{2} - 4ac = 0}}

when the root is real