Question

x square _16 and find the zeros of the following quadratic polynomial and verify the relationship between zeros and the coefficients​

Answer 1

Answer:

4 , -4

Step-by-step explanation:

$given \\ \\ {x}^{2} - 16 = 0 \\ \\ {x}^{2} + 4x - 4x - 16 = 0 \\ \\ x(x + 4) - 4(x + 4) = 0 \\ \\ (x + 4)(x - 4) = 0 \\ \\ x _1 = 4 \\ \\ x_2 = - 4 \\ \\ x_1 + x_2 = 0 \\ \\ x_1 \times x_2 = - 16$

$for \: \: a {x}^{2} + bx + c = 0 \\ \\ if \: \: \alpha \: and \: \beta \: \: are \: roots \: \: then \\ \\ \alpha + \beta = - \frac{b}{a} \\ \\ \alpha \times \beta = \frac{c}{a}$

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Answer 2

Answer:

4 , -4

Step-by-step explanation:

$\begin{lgathered}given \\ \\ {x}^{2} - 16 = 0 \\ \\ {x}^{2} + 4x - 4x - 16 = 0 \\ \\ x(x + 4) - 4(x + 4) = 0 \\ \\ (x + 4)(x - 4) = 0 \\ \\ x _1 = 4 \\ \\ x_2 = - 4 \\ \\ x_1 + x_2 = 0 \\ \\ x_1 \times x_2 = - 16\end{lgathered}$

$\begin{lgathered}for \: \: a {x}^{2} + bx + c = 0 \\ \\ if \: \: \alpha \: and \: \beta \: \: are \: roots \: \: then \\ \\ \alpha + \beta = - \frac{b}{a} \\ \\ \alpha \times \beta = \frac{c}{a}\end{lgathered}$