Question

(3 – x)(2x – 1) > 0 ? Me ajudem pfvr

Answer 1

Resposta:

(3 - x)(2x - 1) > 0

6x - 3 -2x² + x > 0

-2x² + 7x - 3 > 0

-4x² + 14x - 6 > 0

7 ± √25/4

7 ± 5/4

x¹ = 7 + 5/4

x¹ = 12/4

x¹ = 3

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x" = 7 - 5 / 4

x" = 2/4

x" = 1/2

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{x e IR/ x"= 1/2, x'= 3}

Explicação passo-a-passo:

Use ∆ = b² - 4ac

depois

x = -b ± √∆ / 2 . a

Answer 2

(3 – x)(2x – 1) > 0

3(2x-1) -x(2x-1)>0

6x - 3 - 2x² + x > 0

-2x² + 6x + x - 3 > 0

-2x² + 7x - 3 > 0

a = -2 b = 7 c = -3

∆ = b²-4.a.c

∆ = 7²-4.(-2).(-3)

∆= 49-24

∆ = 25

x1/2 = -b±√∆ / 2.a

x1/2 = -7±√25 / 2.(-2)

x1/2 = -7±5/-4

x1 = -7-5/-4

x1 = -12/-4

$x1 = \frac{ - 12}{ - 4} = 3$

$x2 = \frac{ - 7 + 5}{ - 4} = \frac{ - 2 \div 2}{ - 4 \div 2} = \frac{ - 1}{ - 2} = \frac{1}{2}$