Question

puteti sa ma ajutati...
Completati astfel incat polinomul obtinut :
1) sa aiba doua radacini reale simple
2) sa aiba o radacina reala multipla de ordinul 2.
3) sa nu aiba radacini reale.

b) P(x) = __x²-10x+25
c) P(x) = 3x²+__x+1​

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

$\it b)\\ \\ \ 2)\ \ P(x)=x^2-10x+25=(x-5)^2\\ \\ P(x)=0 \Rightarrow (x-5)^2=0 \Rightarrow x_1=x_2=5\\ \\ \\ 1)\ \ P(x)= -3x^2-10x+25=-4x^2+x^2-10x+25=x^2=\\ \\ =(x-5)^2-(2x)^2=(x-5-2x)(x-5+2x)=-(x+5)(3x-5)\\ \\ P(x)=0 \Rightarrow -(x+5)(3x-5)=0 \Rightarrow x_1=-5,\ \ x_2=\dfrac{5}{3}$

$\it 3)\ \ P(x)=2x^2-10x+25=0, \ \ \Delta =-100<0 \Rightarrow x_{1,2}\not\in\mathbb{R}$

$\it c)\\ \\ P(x)=3x^2+bx+1=0\\ \\ \Delta = b^2-12=b^2-(2\sqrt3)^2\\ \\ \\ 3) \ \Delta<0 \Rightarrow b^2<(2\sqrt3)^2 \Rightarrow \sqrt{b^2}<\sqrt{(2\sqrt3)^2} \Rightarrow |b|<2\sqrt3 \Rightarrow\\ \\ \Rightarrow b\in(-2\sqrt3,\ 2\sqrt3);\ \ b=0 \Rightarrow P(x)=3x^2+1=0 \Rightarrow x_{1,2}\not\in\mathbb{R}\\ \\ 2)\ \ \Delta=0 \Rightarrow b^2=(2\sqrt3)^2 \Rightarrow b=\pm2\sqrt3\\ \\ 1) \ \ \Delta>0 \Rightarrow b\in\mathbb{R}\setminus \[[-2\sqrt3,\ \ 2\sqrt3]$

$\it b=4 \Rightarrow P(x)=3x^2+4x+1=3x^2+3x+x+1=3x(x+1)+(x+1)=\\ \\ =(x+1)(3x+1);\ \ P(x)=0 \Rightarrow (x+1)(3x+1)=0 \Rightarrow x_1=-1,\ \ x_2=-\dfrac{1}{3}$