Question

X ao quadrado -3x+5+(x ao quadrado +2x-4)+(x ao quadrado +5x-1)​

Answer 1

    $\Rightarrow \ \ \ x^{2} \ - \ 3x \ + \ 5 \ + \ (x^{2} \ + \ 2x \ - \ 4) + \ (x^{2} \ + \ 5x \ - \ 1) \\\\ \\ \Rightarrow \ \ \ x^{2} \ - \ 3x \ + \ 5 \ + \ x^{2} \ + \ 2x \ - \ 4 \ + \ x^{2} \ + \ 5x \ - \ 1 \\\\ \\ \Rightarrow \ \ \ x^{2} \ + \ x^{2} \ + \ x^{2} \ - \ 3x \ + \ 2x \ + \ 5x \ + \ 5 \ - \ 4 \ - \ 1 \\\\ \\ \Rightarrow \ \ \ 3x^{2} \ + \ 4x \ + \ 0 \\ \\\\ \Rightarrow \ \ \ 3x^{2} \ + \ 4x \ = \ 0$

•  $a \ = \ 3 \ \ ; \ \ \ \ \ b \ = \ 4 \ \ ; \ \ \ \ \ c \ = \ 0$

    $\Delta \ = \ b^{2} \ - \ 4ac \\\\ \Delta \ = \ 4^{2} \ - \ 4 \ . \ 3 \ . \ 0 \\\\ \Delta \ = \ 16$

    $x \ = \ \frac{-b \ \ \± \ \ \sqrt{\Delta}}{2a} \\\\ x \ = \ \frac{-4 \ \ \± \ \ \sqrt{16}}{2 \ . \ 3} \\\\ x \ = \ \frac{-4 \ \ \± \ \ 4}{6}$

•  $x' \ = \ \frac{-4 \ + \ 4}{6} \ \ \ \ \ \Rightarrow \ \ \ \ \ x' \ = \ \frac{0}{6} \ \ \ \ \ \Rightarrow \ \ \ \ \ \boxed{ \ x' \ = \ 0 \ }$
• $x'' \ = \ \frac{-4-4}{6} \ \ \ \ \ \Rightarrow \ \ \ \ \ x'' \ = \ -\frac{8}{6} \ \ \ \ \ \Rightarrow \ \ \ \ \ \boxed{ \ x'' \ = \ - \frac{4}{3} \ }$