Question

Fie A = $x^{2}$ - 2x - 15
B = $x^{2}$ + 6x + 9
C = $x^{2}$ - 10x + 25
a. Demonstrati ca $A^{2}$ = BC
b. Simplificati raportul $\frac{C+A}{C-A}$

Answer 1

$A=x^2-5x+3x-15=x(x-5)+3(x-5)=(x+3)(x-5). \\ \\ B=(x+3)^2 \\ \\ C=(x-5)^2 \\ \\ a)~A^2=[(x+3)(x-5)]^2=(x+3)^2 \cdot (x-5) ^2=BC. \\ \\ b)~ \frac{C+A}{C-A}= \frac{(x-5)^2+(x+3)(x-5)}{(x-5)^2-(x+3)(x-5)} = \frac{(x-5)(x-5+x+3)}{(x-5)(x-5-x-3)}= \frac{2x-2}{-8}= \frac{1-x}{4}.$

Answer 2

a)
$B = x^2+6x+9=x^2+2*x*3+3^2=(x+3)^2 \\ C = x^2-10x+25=x^2-2*x*5+5^2=(x-5)^2$
Descompunem in factori pe  - 2x - 15:
$x^2-2x-15=x^2-5x+3x-15=x(x-5)+3(x-5)=\\=(x-5)(x+3)$
Atunci:
A=(x-5)(x+3) => A²=[(x-5)(x+3)]²
B * C = (x+3)²(x-5)²=[(x-5)(x+3)]²
=>A²=B*C
b) $\frac{C+A}{C-A}=\frac{(x-5)^2+(x-5)(x+3)}{(x-5)^2-(x-5)(x+3)}=\frac{(x-5)(x-5+x+3)}{(x-5)(x-5-x-3)} = \\ = \frac{2x-2}{-8}=\frac{2(x-1)}{-8}=\ \textgreater \ \frac{C+A}{C-A}=\frac{x-1}{-4}=-\frac{x-1}{4}=\frac{1-x}{4}$