Question

The answer to the first question.

Answer 1

Answer:

see explanation

Step-by-step explanation:

Given

$5x^2 - 4 - 8x$

To find the zeros equate to zero and rearrange into standard form, that is

$5x^2 - 8x - 4 = 0$ ← in standard form

To factor the quadratic

Consider the factors of the product of the coefficient of the x² term and the constant term which sum to give the coefficient of the x- term.

$\text{Product} = 5 \times - 4 = - 20 \text{ and Sum}= - 8$

The factors are - 10 and + 2

Use these factors to split the x- term

$5x^2 - 10x + 2x - 4 = 0$ ( factor the first/second and third/fourth terms )

$5x(x - 2) + 2(x - 2) = 0$ ← factor out (x - 2) from each term

$(x - 2)(5x + 2) = 0$

Equate each factor to zero and solve for x

$x - 2 = 0 \implies x = 2\\5x + 2 = 0 \implies 5x = - 2 \implies x = -\frac{2}{5}$

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$\text{The sum of the zeros} = - \frac{b}{a}$

$\text{The product of the zeros} = \frac{c}{a}$

$\text{with } a = 5,\: b = - 8 \text{ and } c = - 4$

$\text{The sum} = 2 - \frac{2}{5}= \frac{8}{5}$

$\text{and} - \frac{b}{a} = - \frac{-8}{5}= \frac{8}{5}$

Thus verified

$\text{The product} = 2 \times -\frac{2}{5}= - \frac{4}{5}$

$\text{and } \frac{c}{a} = \frac{-4}{5}= - \frac{4}{5}$

Thus verified.