For what values of m does f(x)=mx^2 - 2x +1 have no real roots?

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answers:
Como say: For real roots :-
b² ≥ 4 a c
4 ≥ 4 m
1 ≥ m
No real roots for m > 1
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lenpol7 say: Apply the discriminant
b^2 - 4ac < 0
(-2)^2 - 4(m)(1) < 0
4 - 4m < 0
4 < 4m
4m > 4
m > 1
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la console say: mx² - 2x + 1 = 0

Polynomial like: ax² + bx + c, where:
a = m
b = - 2
c = 1

Δ = b² - 4ac (discriminant)

Δ = (- 2)² - 4.(m * 1)

Δ = 4 - 4m → When Δ < 0, there is no root

4 - 4m < 0

- 4m < - 4

4m > 4

m > 1
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say: we need the discriminant
so b^2-4ac
= (-2)^2-4(m)(1)
= 4-4(m)
= 4-4m
for "no real roots"
the discriminant < 0
so 4-4m<0
so m>1
hope this helps
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ohevl say: saauuuki
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farr say: fdfd
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