How do you solve $4x^2-64=0$ using the quadratic formula?

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Answer 1 · 2017-08-19T15:03:50

See a solution process below:

We can rewrite the equation as:

$4x^2 + 0x - 64 = 0$

For $color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0$ , the values of $x$ which are the solutions to the equation are given by:

$x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))$

Substituting:

$color(red)(4)$ for $color(red)(a)$

$color(blue)(0)$ for $color(blue)(b)$

$color(green)(-64)$ for $color(green)(c)$ gives:

$x = (-color(blue)(0) +- sqrt(color(blue)(0)^2 - (4 * color(red)(4) * color(green)(-64))))/(2 * color(red)(4))$

$x = +- sqrt(0 - (-1024))/8$

$x = +- sqrt(1024)/8$

$x = +- 32/8$

$x = +- 4$

How do you solve 4x^2-64=04x^2-64=0 using the quadratic formula?