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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​

User Jon Nagra
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1 Answer

1 vote

Answer:

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.

User Jackee
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