Question

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​

Answer 1

m<1

Explanation:

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

`b^2-4ac>0\cdots(i)`.

The given expression is
`x^2-2x+m`.

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

So, by using equation (i),

`(-2)^2-4(1)(m)>0`

`\Rightarrow 4-4m>0`

`\Rightarrow 4m<4`

`\Rightarrow m<4/4`

`\Rightarrow m<1`.

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