Question #7c2d2

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Answer 1 · 2018-01-25T04:52:30

$x=+-4/3$

$9x^2-16=0$

Using the difference of squares, $a^2-b^2=(a-b)(a+b)$

$:.(3x-4)(3x+4)=0$

We get two solutions:

$3x-4=0$ , $3x=4$ , $x=4/3$

$3x+4=0$ , $3x=-4$ , $x=-4/3$

Combining the two solutions, we can write

$x=+-4/3$

Answer 2 · 2018-01-25T05:02:13

$9x^2-16=0$

Let's start by adding $color(red)(16)$ to both sides

$9x^2 cancel(-16)cancelcolor(red)(+16) = 0 + color(red)(16)$

$9x^2 = 16$

Since we are solving for $x$ , divide both sides by $9$

$(cancel9x^2)/cancel9 = 16/9$

$x^2=16/9$

Now take square root

$x=+-sqrt(16/9)$

Simplify it

$x= +-4/3$