For what values of m are there 2 different distinct real solutions? Give your answer as an inequality. X^2-2x+m.

I'm not sure, but I think the formula is b^2-4ac>0​

Respuesta :

Answer:

m<1

Step-by-step explanation:

If the expression is in the form of [tex]ax^2+bx+c[/tex], then the condition for having 2 different real roots is

[tex]b^2-4ac>0\cdots(i)[/tex].

The given expression is [tex]x^2-2x+m[/tex].

Here, a=1, b=-2 and c=m.

So, by using equation (i),

[tex](-2)^2-4(1)(m)>0[/tex]

[tex]\Rightarrow 4-4m>0[/tex]

[tex]\Rightarrow 4m<4[/tex]

[tex]\Rightarrow m<4/4[/tex]

[tex]\Rightarrow m<1[/tex].

Hence, the condition for the expression [tex]x^2-2x+m[/tex] to have two different real roots is m<1.