FUNKCJA KWADRATOWA – rozwiązania zadań

Zadanie 1

(m+1)x² − 4mx + m + 1 = 0

a = m + 1

b = −4m

c = m + 1

1    m + 1 ≠ 0  ⇒ m ≠ −1

2    > 0

3 x₁ * x₂ > 0

4 x₁ + x₂ > 0

Ad.1

m ≠ −1

Ad.2

b² − 4ac > 0

16m² − 4(m+1)(m+1) > 0

16m² − 4(m²+2m+1) > 0

16m² − 4m² − 8m − 4 > 0

12m² − 8m − 4 > 0

3m² − 2m − 1 > 0

=16

$$\sqrt{} = 4$$

$$m_{1} = - \frac{1}{3}$$

m₂ = 1

$$m \in \left( - \infty, - \frac{1}{3} \right)(1, + \infty)$$

Ad.3

$$x_{1}*x_{2} < 0\ \Leftrightarrow \frac{c}{a} < 0$$

$$\frac{m + 1}{m + 1} > 0$$

1 > 0

m₁ = −1

m₂ = −1

m₁ = m₂

m ∈ R

Ad.4

$$x_{1} + x_{2} > 0\ \Leftrightarrow - \frac{b}{a} > 0$$

$$\frac{4m}{m + 1} > 0\ \ \ \ \ \ \ \ \ \ \ \ |\left( m + 1 \right)^{2}$$

4m(m+1) > 0

4m = 0  ⇒ m = 0

m + 1 = 0  ⇒ m = −1

m₁ = 0,  m₂ = −1

m ∈ (−∞,−1) ∪ (0, ∞)

Odpowiedz :    m ∈ (−∞,−1) ∪ (1, +∞)

Zadanie 2

x² − (m−5)x + 2(3−m) = 0

1   a ≠ 0  ⇒ 1 ≠ 0 dla m ∈ R

2    ≥ 0

b² − 4ac ≥ 0

(m−5)² − 4 * 2(3−m) ≥ 0

m² − 10m + 25 − 24 + 8m ≥ 0

m² − 2m + 1 ≥ 0

=4 − 4 = 0

$$m = \frac{- b}{2a}$$

$$m = \frac{2}{2} = 1$$

m ∈ R

$$\mathbf{3}\text{\ \ \ }\frac{1}{x_{1}} + \frac{1}{x_{2}} < 0$$

$$\frac{x_{2} + x_{1}}{x_{1}x_{2}} < 0$$

$$\frac{- \frac{b}{a}}{\frac{c}{a}} < 0$$

$$- \frac{b}{c} < 0$$

$$\frac{m - 5}{2\left( 3 - m \right)} < 0\ \ \ \ \ |\ {2\left( 3 - m \right)}^{2}$$

(m−5)(3−m) < 0

m = 5

m = 3

Odpowiedz :     m ∈ (−∞,3) ∪ (5, +∞)

Zadanie 3

(m−1)x² − 2mx + m − 2 = 0

m − 1 = 0  ⇒  m ≠ 1

a = m − 1

b = −2m

c = m − 2

1    > 0

>0 ⇔ b² − 4ac > 0

4m² − 4(m−1)(m−2) > 0

4m² − 4(m²−2m−m+2) > 0

4m² − 4m² + 8m + 4m − 8 > 0

12m > 8

$$m > \frac{2}{3}$$

$$m \in (\frac{2}{3}, + \infty)$$

2   x₁x₂ > 0

$$x_{1}x_{2} > 0\ \Leftrightarrow \frac{c}{a} > 0$$

$$\frac{m - 2}{m - 1} > 0$$

m = 2

m = 1

m ∈ (−∞,1) ∪ (2, +∞)

3 x₁ + x₂ < 0

$$x_{1} + x_{2} < 0\ \Leftrightarrow \ - \frac{b}{a} < 0$$

$$\frac{2m}{m - 1} < 0$$

2m(m−1) > 0

m = 0                  m = 1

m ∈ (0, 1)

Zadanie 4

(1+m)x² + 3x + 7 > 0

a = 1 + m

b = 3

c = 7

1   a > 0

1 + m > 0

m > −1

m ∈ (−1,+∞)

2    < 0

b² − 4ac < 0

9 − 4(1+m) * 7 < 0

9 − 28 − 28m < 0

−28m < 19

$$m > - \frac{19}{28}$$

$$Odpowiedz:\ \ \ m \in ( - \frac{19}{28}, + \infty)$$

Zadanie 5

mx² + 2(m+1)x + m = 0

I sposob

a = m

b = 2(m+1)

c = m

x₁² + x₂² > x₁ + x₂

$$\left( x_{1} + x_{2} \right)^{2} - 2x_{1}x_{2} > - \frac{b}{a}$$

$$\left( - \frac{b}{a} \right)^{2} - \frac{2c}{a} > - \frac{b}{a}$$

$$\frac{b^{2}}{a^{2}} - \frac{2c}{a} + \frac{b}{a} > 0$$

$$\frac{b^{2} - 2ac + ab}{a^{2}} > 0$$

$$\frac{4\left( m + 1 \right)^{2} - 2m^{2} + 2m\left( m + 1 \right)}{m^{2}} > 0$$

$$\frac{4\left( m^{2} + 2m + 1 \right) - 2m^{2} + 2m^{2} + 2m}{m^{2}} > 0$$

$$\frac{4m^{2} + 8m + 4 + 2m}{m^{2}} > 0$$

$$\frac{4m^{2} + 10m + 4}{m^{2}} > 0$$

4m² + 10m + 4 = 0

=100 − 16 * 4 = 100 − 64 = 36

$$\sqrt{} = 6$$

$$x_{1} = \frac{- 10 - 6}{8} = - 2$$

$$x_{2} = \frac{- 10 + 6}{8} = - \frac{1}{2}$$

m² = 0

m = 0

$$Odpowiedz:\ \ \ m \in \left( - \frac{1}{2},0 \right) \cup (0, + \infty)$$

II sposob

1   a ≠ 0

a = m ⇒ m ≠ 0

2    ≥ 0

$$m \geq - \frac{1}{2}$$

zestawienie zalozen

$$Odpowiedz:\ \ \ \ m \in \left( - \frac{1}{2},0 \right) \cup (0, + \infty)$$