Question

The quadratic polynomial x^2-4x+k=0 has distinct real roots

Answer 1

Question :

The quadratic polynamial x²-4x+k = 0 has distinct real roots.

AnsWer :

K<4.

Required Information :

$\blacksquare \tt \: D = {b}^{2} - 4ac. \\ \blacksquare \tt \: D \: \: \: means \: Discriminant .\\ \blacksquare\tt \: distinct \: real \: roots \: means \\ \: \: \: \tt \: D > 0.$

To Find :

Value of k.

Solution :

we have quadratic Equation,

$\: \tt {x}^{2} - 4x + k = 0. \\ \tt where \: as \: a = 1. \: \: b = - 4. \: \: \: c = k.$

Now,

$\tt \: D = {b}^{2} - 4ac. \\ \: \: \: \: \tt= { (- 4)}^{2} - 4 \times 1 \times k. \\ \: \: \: \: \tt = 16 - 4k.$

But, it have distinct roots.

$\: \: \: \: \: \tt \: D > 0. \\ \leadsto \tt{b}^{2} - 4ac > 0.$

$\leadsto \tt16 - 4k > 0. \\ \leadsto \tt16 > 4k. \\ \leadsto \tt4 > k. \\ \leadsto \tt k < 4.$

Therefore, the value of distinct roots of Equation be k < 4.