Question

Sa se rezolve sistemul de ecuatii.
a)2x+1=y
y=X^2-2x+5

b) y=2x-1
x^2-4x-5=y

Answer 1

Răspuns:

a) 2x+1=y

y=x²-2x+5

=> 2x+1=x²-2x+5

2x-x²+2x = 5-1

4x-x² =4

4x-x²-4 =0

-(x²-4x-4)=0

-(x-2)² =0

(x-2)² =0

x-2=0

x=0+2

x=2

=> y= 2²- 2×2+5

y= 4-4+5

y=0+5

y=5

b)y=2x-1

y=x²-4x-5

=> 2x-1 = x²-4x-5

2x-1-x²+4x+5=0

6x+4-x²=0

-x²+6x+4=0 a=-1 b=6 c=4

delta = b²-4ac = 6² -4×(-1)×4 =36+16=52

$x1 = \frac{ - b + \sqrt{delta} }{2a} = \frac{ - 6 + \sqrt{52} }{2 \times ( - 1)} = \frac{ - 6 + 2 \sqrt{13} }{2( - 1)} = - \frac{2( - 3 + \sqrt{13}) }{2} = - ( - 3 + \sqrt{13} = 3 - \sqrt{13}$

$x2 = \frac{ - b - \sqrt{delta} }{2a} = \frac{ - 6 - \sqrt{52} }{2 \times ( - 1)} = \frac{ - 6 - 2 \sqrt{13} }{2 \times ( - 1)} = - \frac{2( - 3 - \sqrt{13} }{2} = - ( - 3 - \sqrt{13} ) = 3 + \sqrt{13}$

$y1 = 2 \times (3 - \sqrt{13} ) - 1 = 6 - 2 \sqrt{13} - 1 = 5 - 2 \sqrt{13} \\ y2 = 2 \times (3 + \sqrt{13} ) - 1 = 6 + 2 \sqrt{13} - 1 = 5 + 2 \sqrt{13}$

Sper ca te-am ajutat :)