Question

rozwiąż równania:

(3x-8)² - (4x-6)²=28

x⁴+13x²-48=0

(5x-4)²-(3x+1)=15

x⁴+7x²-18=0

Δ=b²-4ac tym wzorem prosze o dokladne obliczenia

Answer 1

1.

(3x-8)^2 -(4x-6)^2 = 28

9x^2 - 48x + 64 -(16x^2 - 48x +36) = 28

9x^2 - 48x + 64-16x^2 + 48x - 36 = 28

-7x^2 = 0

x = 0

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2.

x^4 +13x^2 - 48 = 0

x^2 = t 

 

t^2 +13t - 48 = 0

Dt = b^2 - 4ac

Dt = 13^2 - 4*(-48) = 361

VDt = 19

t1 =(-b+VD)/2a = (-13-19)/2 = -16  --odpada,bo x^2 >= 0

t2 = (-13+19)/2 = 3 

 

x2 = 3 

x^2 -3 = 0

(x + V3)(x- V3) = 0

x = -V3   v   x = V3

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3.

(5x-4)^2 -(3x+1) = 15

25x^2 - 40x =16 -3x -1 = 15

25x^2 - 43x = 0

x(25x - 43) = 0

x = 0

lub

25x = 43

x = 43/25 = 1 18/25

x = 0   v   x = 1 18/25

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4.

x^4 + 7x2 -18 = 0

x^2 = t

 

t^2 + 7t -18 = 0

t = 49 - 4 *(-18) = 121

VD = 11

t1 = (-7-11)/2 = -9   -- odpada,bo x^2 >= 0

t2 = (-7+11)/2 = 2 

 

x^2 = 2 

x^2 - 2 = 0

(x+V2)(x-V2) = 0

x = -V2   v   x = V2

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