Question

1) rationalizarea fractiei 4 supra √5 -1
2) Solutia negativa a ecuatiei (x+1)²=16

Answer 1

[tex]1. \frac{4}{ \sqrt{5}-1}= \frac{4( \sqrt{5}+1)}{( \sqrt{5}-1)(\sqrt{5}+1)}= \frac{4( \sqrt{5}+1)}{\sqrt{5} ^{2}-1^{2}}= \frac{4( \sqrt{5}+1)}{5-1}= \frac{4( \sqrt{5}+1)}{4}= \sqrt{5}+1 2.(x+1)^{2}=16 x^{2}+2x+1=16 x^{2}+2x-15=0 [/tex]

Δ=b²-4ac=4-4*(-15)=4+60=64

x1= (-b+√Δ)/2a

x2=(-b-√Δ)2a= (-2-√64)/2=(-2-8)/2= -10/2= -5

Answer 2

2)...  (x+1)²=16  ⇔  x+1 = $|\pm4|$ 
         ⇒  solutiile pot fi:  x = {+4-1=+3 ; -4-1=-5}  din care solutia negativa este: -5  ;
1)...  pentru rationalizarea fractiei , amplificam fractiea cu: √5+1 !
        adica:  [tex]\frac{^{\sqrt5+1)}4}{\sqrt5-1}=\frac{4\sqrt5+4}{5-1}=\\ =\frac{\not4(\sqrt5+1)}{\not4}=\sqrt5+1[/tex]