Question

When solving a problem, two students fell back on a 2nd degree equation, however, when solving it, they both made a mistake. The 1st student only got the independent term of the equation wrong, finding the roots -10 and -2. The 2nd student only got the coefficient of the 1st degree term wrong, finding roots 4 and 8. Were the correct roots of the equation?

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Answer to a math question When solving a problem, two students fell back on a 2nd degree equation, however, when solving it, they both made a mistake. The 1st student only got the independent term of the equation wrong, finding the roots -10 and -2. The 2nd student only got the coefficient of the 1st degree term wrong, finding roots 4 and 8. Were the correct roots of the equation?

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Santino
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Com certeza, vamos descobrir as raízes corretas da equação quadrática! Veja como podemos abordar isso: **Compreendendo as informações** * **Aluno 1:** Errou no termo constante (termo independente). Suas raízes são -10 e -2. * **Aluno 2:** Errou no coeficiente do termo x. Suas raízes são 4 e 8. **Usando as propriedades das raízes** 1. **Soma das Raízes:** Em uma equação quadrática *ax² + bx + c = 0*, a soma das raízes é igual a *-b/a*. 2. **Produto das Raízes:** Em uma equação quadrática *ax² + bx + c = 0*, o produto das raízes é igual a *c/a*. **Vamos analisar:** * **Aluno 1:** * Produto de suas raízes: (-10) * (-2) = 20. Este pode ser o *c/a* correto. * **Aluno 2:** * Soma de suas raízes: 4 + 8 = 12. Este poderia ser o *-b/a* correto. **Encontrando a equação correta:** Vamos supor que a equação quadrática correta seja *x² + bx + c = 0*. (Definiremos a = 1 para simplificar). * Do Aluno 2, temos -b/1 = 12, então b = -12 * Do Aluno 1, temos c/1 = 20, então c = 20 **A equação correta:** x² - 12x + 20 = 0 **Encontrando as raízes corretas** Podemos resolver esta equação quadrática usando fatoração ou a fórmula quadrática: (x - 10) (x - 2) = 0 Portanto, as raízes corretas são x = 10 e x = 2. **Conclusão:** As raízes corretas da equação eram 10 e 2.

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